Quadratic Equations

(i) On solving the equations,

Re-writing in the format of ax² + bx + c = 0

∵ b = 0 & c = 2

So, following the ideal pattern of a quadratic polynomial is a quadratic polynomial.

(ii) On solving the equations,

x+

it can’t be re-written in the format of ax² + bx + c = 0

So, following the ideal pattern of a quadratic polynomial, x+ is not a quadratic polynomial.

(iii) On solving the equations,

x+

∵ it can’t be re-written in the format of ax² + bx + c = 0

So, following the ideal pattern of a quadratic polynomial, x+is not a quadratic polynomial.

(iv) On solving the equations,

x² + 3√x + 2

it can’t be re-written in the format of ax² + bx + c = 0

So, following the ideal pattern of a quadratic polynomial, x² + 3√x + 2 is not a quadratic polynomial.

(i) On solving the equations,

2x² + 1 = 0

Re-writing in the format of ax² + bx + c = 0

(2)x² + (0)x + 1 = 0

a = 2 b = 0 & c = 1.

So, following the ideal pattern of a quadratic polynomial 2x² + 1 is a quadratic polynomial.

(ii) On solving the equations,

it can’t be re-written in the format of ax² + bx + c = 0

So, following the ideal pattern of a quadratic polynomial, x²+ is not a quadratic polynomial.

(iii) On solving the equations,

it can’t be re-written in the format of ax² + bx + c = 0

So, following the ideal pattern of a quadratic polynomial, is not a quadratic polynomial.

(iv) On solving the equations,

3√(x²+ 1) + x = 0

it can’t be re-written in the format of ax² + bx + c = 0

So, following the ideal pattern of a quadratic polynomial, 3√(x²+ 1) + x = 0 is not a quadratic polynomial.

(i) On solving the equations,

After simplifying the equation, one of the term has a negative (-2) exponent.

So, following the ideal pattern of a polynomial, is not a polynomial.

(ii) On solving the equations,

∵ After simplifying the equation, as it has a positive (2) exponent.

So, following the ideal pattern of a polynomial, is a polynomial.

(iii) On solving the equations,

∵ After simplifying the equation, as the one of the term has a negative (-3) exponent.

So, following the ideal pattern of a polynomial, is not a polynomial.

(iv) On solving the equations,

After simplifying the equation, as the expression has a degree of .

So, following the ideal pattern of a polynomial, is not a polynomial.

(i) Quadratic, because it is in the form of ax² + bx² + c = 0

(ii) n = 2, and also a≠0, as it will make the polynomial 0.

(iii) Putting the value of x = -1, in x² — 5x + 4

(-1)² - 5(-1) + 4

1 + 5 + 4

10

The value is 10.

(iv) The degree is the highest power of the term in the expression, so it is 3.

(v) ∵ The zeroes of a polynomial are α & β.

∴ to be zero, αx²+b α+c=0 & βx²+b β+c=0

-x² + 9 = 0

-x² = -9

x² = 9

∴ x = ±3

∴ The zeroes of the given polynomial are 3 & -3.

4x² - 1 = 0

4x² = 1

∴

The zeroes of the given polynomial are & .

9 - 4x² = 0

4x² = 9

∴

The zeroes of the given polynomial are & and the option (c) is correct.

x² =8

∴ x = ±2√2

The zeroes of the given polynomial 2√2 and the option (b) is correct.

Putting the value of -2, in the given polynomial,

3(-2)² + (-2) – 10

3(4) – 2 – 10

12 – 12

0

∵ the value comes out to be 0.

∴ -2 is one of the zeroes and, yes 3x² + x — 10 is a quadratic polynomial.

Putting the value of -1, in the given polynomial,

(-1)² + 2(-1) – 3

1 – 2 – 3

3 – 3

0

∵ the value comes out to be 0.

∴ -1 is one of the zeroes of the given polynomial.

The highest power is 2, so the degree is also 2.

Equating the expression with 0,

x² = 4

∴ x = ±2

Yes, the above expression is a polynomial, as it has no negative powers in any of the terms and its zeroes are 2 & -2.

∵ the power of a term is in negative ().

∴ The above given expression is not a polynomial.

In the above expressions, only the thirdone has the positive power unlike others.

∴ It is the only polynomial.

Equating the expression with 0,

x² = 16

∴ x = ±4

The zeroes of the polynomial are 4 & -4.

In the above expressions, only the secondone has a positive power, unlike others.

∴ It is the only polynomial.

Equating the expression with 0,

x² + 2x = 0

x ( x+2) = 0

∴ x = 0 or x + 2 = 0

X = 0 Or x = -2

The zeroes of the polynomial x² +2x are 2 & -2, having a degree of 2, being the highest power of the terms in the same expression.

∵ The highest power is 2, so the degree is also 2, in both the expressions.

Equating the expression with 0,

x² =16

∴ x = ±4

Yes, the above expression (⁾ is a polynomial, and its zeroes are 4 & -4.

Equating the expression with 0,

z² + z + 1 =0

Using Sreedharacharya formula,

ax²+bx+c = 0

x =

x =

x =

∵ it does not have real values.

∴ The zeroes of z² + z + 1 are complex numbers, though it is a polynomial having the degree 2.

Equating the expression with 0,

x² — 6x + 8 = 0

On factorising it further,

x² - 4x - 2x + 8 = 0

x(x - 4) – 2(x - 4) = 0

(x - 4) (x - 2) = 0

∴ x = 4 or x = 2

∴ The zeroes of x² — 6x + 8 are 4 & 2.

Equating the expression with 0,

2x² +x -1 = 0

On factorising it further,

2x² -x + 2x - 1 = 0

x(2x - 1) +1(2x - 1) = 0

(2x - 1) (x + 1) = 0

x = or x = -1

∴ The zeroes of 2x² + x — 1 are and -1.

Equating the expression with 0,

2x² - 5x + 2 = 0

On factorising it further,

2x² - 4x- x + 2 = 0

2x(x - 2) -1(x - 2) = 0

(2x - 1) (x - 2) = 0

x = or x = 2

The zeroes of 2x²— 5x + 2 are and 2.

Equating the expression with 0,

5x² - 4x - 1 = 0

On factorising it further,

5x² - 5x + x - 1 = 0

5x(x - 1) +1(x - 1)=0

(5x + 1) (x - 1) = 0

x = - or x = 1

The zeroes of 5x² - 4x — 1 are - and 1.

Equating the expression with 0,

x² — 2x + 3= 0

Using Sreedharacharya formula,

ax²+bx+c = 0

x =

x =

x =

it does not have real values.

The zeroes of x² — 2x + 3 are complex numbers.

Equating the expression with 0,

3x² - 10x + 3 = 0

On factorising it further,

3x² - 9x - x + 3 = 0

3x(x - 3) - 1(x - 3) = 0

(3x - 1) (x - 3) = 0

x = or x = 3

The zeroes of 3x² — 10x + 3 are and 3.

Equating the expression with 0,

3x² + 5x + 2 = 0

On factorising it further,

3x² + 3x + 2x + 2 = 0

3x(x + 1) + 2(x + 1) = 0

(3x + 2) (x + 1) = 0

x = - or x = -1

The zeroes of 3x² + 5x + 2 are – and -1.

Equating the expression with 0,

4x² — x — 5 = 0

On factorising it further,

4x² + 4x - 5x - 5 = 0

4x(x + 1) -5(x + 1)=0

(4x - 5) (x + 1) = 0

x = or x = -1

The zeroes of 4x² — x — 5 are and -1.

Given; (x − 2) (x + 1) = (x − 1) (x + 3)

⇒ x² + x − 2x − 2 = x² + 3x − x − 3

⇒ x² − x − 2 − x² − 2x + 3 = 0

⇒ −3x + 1 = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given; (x − 2)² + 1 = 2x − 3

⇒ x² − 2x + 4 + 1 = 2x − 3

⇒ x² − 2x + 5 − 2x + 3 = 0

⇒ x² − 4x + 8 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

Given; x (x + 1) + 8 = (x + 2) (x − 2)

⇒ x² + x + 8 = x² − 2²

⇒ x² + x + 8 − x² + 4 = 0

⇒ x + 12 = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given; (x − 3) (2x + 1) = x (x + 2)

⇒ 2x² + x − 6x − 3 = x² + 5x

⇒ 2x² + x − 6x − 3 − x² − 5x = 0

⇒ x² − 10x − 3 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

Given; x (2x + 3) = x² + 1

⇒ 2x² + 3x = x² + 1

⇒ 2x² + 3x − x² − 1 = 0

⇒ x² +3x − 1 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

Given; x² + 3x + 1 = (x − 2)²

⇒ x² + 3x + 1 = x² − 4x + 4

⇒ x² + 3x + 1 − x² − 4x − 4 = 0

⇒ −x − 3 = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given; (x + 1) (x − 1) = (x + 2) (x + 3)

⇒ x² − 1² = x² + 5x + 6

⇒ x² − 1 − x² − 5x − 6 = 0

⇒ −5x − 7 = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given; (x − 1)² = (x + 1)²

⇒ x² − x + 1² = x² + x + 1²

⇒ x² − x + 1 − x² − x − 1 = 0

⇒ −2x = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given; (x + 2)³ = x³ − 4

⇒ x³ + 6x² + 12x + 2³ = x³ − 4

⇒ x³ + 6x² + 12x + 2³− x³ + 4 = 0

⇒ 6x² + 12x + 12 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

Given;

⇒ x² − 1 = 8x

⇒ x² − 8x − 1 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

Given;

⇒

⇒

⇒ 4x⁴ + 20x² + 25 = 9x

⇒ 4x⁴ + 20x² − 9x + 25 = 0

∵ The highest power of x in the equation is 4;

∴ It is not a quadratic equation.

Given;

⇒ x³ + 1 = 5x

⇒ x³− 5x + 1 = 0

∵ The highest power of x in the equation is 3;

∴ It is not a quadratic equation.

Given;

⇒ x⁴− 1 = 8x²

⇒ x⁴− 8x²− 1 = 0

∵ The highest power of x in the equation is 4;

∴ It is not a quadratic equation.

Let ‘x’ be the number of marbles John had.

∴ Jivanti will have 45 − x marbles.

[∵ the total number of marbles is 45]

As both of them lost 5 marbles each; marbles they now have will be x − 5 and 40 − x respectively.

⇒ Product of the number of marbles = (x − 5) (40 − x)

∴ 40x − x² − 200 + 5x = 124 [∵ the product is 124]

⇒ x² − 45x + 324 = 0

Let ‘x’ be the number of books bought by the shopkeeper.

∴ Cost of one book = Rs. 80 ÷ x.

[∵ the total cost of books is Rs. 80]

∴ Cost of one book when he buys 4 more books for same rate = Rs. 80 ÷ (x + 4).

When he buys four more books for the same amount; it will cost Re. 1 per book less than the previous.

∴ 80 ÷ (x + 4) = (80 ÷ x) − 1

⇒

⇒ 80x = (x + 4) (80 − x)

⇒ 80x = 80x − x² + 320 − 4x

⇒ x² + 4x + 320 = 0

Let ‘x’ be the number of toys produced on that day.

∴ Cost of production of each toy that day= Rs. 55 − x.

[∵ The cost of production of each toy (in rupees) is 55 minus the number of toys produced in a day]

∴ Cost of production of x toys = Rs. x (55 − x)

[∵ On a particular day, the total cost of production was Rs. 750.]

∴ x (55 − x) = 750

⇒ x² − 55x + 750 = 0

Let ‘x’ and ‘y’ be the smaller and larger integer respectively.

∴ x² + y² = 117.

[∵ The sum of the squares of two positive integers is 117]

∴ x² = 4y

[∵ the square of the smaller number equals four times the larger number.]

∴ y² + 4y – 117 = 0

Let ‘x’ be the one part and the other part will be 16 − x.

[∵ 16 is being divided]

∵ Twice of the square of larger part exceeds the square of the smaller part by 164.

∴ 2x² = (16 – x)² + 164

∴ 2x² = 256 – 32x + x² + 164

⇒ x² + 32x − 420 = 0

Let ‘y’ and ‘x’ be the present age of the man and son.

∵ One year ago the man was eight times as old as his son.

∴ y – 1 = 8 (x − 1)

∵ Now, his age is equal to the square of his son's age.

∴ y = x²

∴ x² − 1 = 8 (x − 1)

⇒ x² − 8x + 7 = 0

Let ‘x’ be the speed of the train in km per hour.

∴ Time taken to cover 300 km =

[∵ .]

∴ Time taken when speed is increased by 5 =

∵ The journey would have taken two hours less when speed is decreased.

Let ‘x’ be the length of the shortest side.

∵ The hypotenuse of a right−angled triangle is 6 metres more than twice of the shortest side.

∴ length of hypotenuse = 2x + 6

∵ The third side is two metres less than the hypotenuse.

∴ length of third side = hypotenuse – 4 = 2x + 4

By applying Pythagoras theorem; Hypotenuse square is equal to sum of the squares of other two sides.

⇒ (2x + 6)² = x² + (2x + 4)²

Let ‘x’ be the length of cloth.

∵ Cost of x metre is Rs. 200.

∴ Cost per metre = Rs. 200 ÷ x

∴ Cost per metre when total size is x + 5 = Rs. 200 ÷ (x + 5)

∵ Cost of 5 metre longer cloth is Rs. 2 less for each metre.

Put both the values of x in the equation.

When

= 0

When

= 0

R.H.S = L.H.S

Therefore, and are the solutions of the given equation.

Put both the values of x in the equation.

When x = 1

1² – 5(1) + 4

1 – 5 + 4

= 0

Therefore, it is the solution to the equation.

When x = 3

3² – 5(3) + 4 = 0

9 – 15 + 4

= –2

Therefore, it is not the solution to the equation.

Put both the values of x in the equation.

When x = √3

(√3)² –3√3(√3) + 6 = 0

3 – (3)3 + 6

3 – 9 + 6

9 – 9

= 0

Therefore, it is the solution to the equation.

When x = –2√3

(–2√3)² –3√3(–2√3) + 6 = 0

12 + 18 + 6

= 36

Therefore, it is not the solution to the equation.

The two possible solutions are x = 3 and .

Since this question is given in standard form, meaning that it follows the form: ax² + by + c = 0, we can use the quadratic formula to solve for x:

x = 3

That value of x is correct as well!

Therefore, the two possible solutions are:

x=3

x=−0.50

Put x = √2

(√2)² + √2(√2) – 4 = 0

2 + 2 – 4

4 – 4

= 0

Therefore, it is the solution to the equation.

When x = –2√2

(–2√2)² + √2(–2√2) – 4 = 0

8 – 4 – 4

= 0

Therefore, it is the solution to the equation.

Put x = –3 in the equation.

(–3)² + 6(–3) + 9

9 – 18 + 9

=0

Hence it is a solution

Put x = –3 in the equation.

2(–3)² + 5(–3) –3 = 0

18 – 15 –3

18 – 18

= 0

Therefore, x = –3 is the solution of the equation.

The given quadratic equation is 3x² + 13x + 14 = 0

Putting x = – 2,

L.H.S.

3.(–2)² + 13.(–2) + 14

3 x 4 – 26 + 14

12 – 26 + 14

26 – 26

= 0

Hence, x = – 2 is a solution of 3x + 13x + 14 = 0

kx² – x – 2 = 0

4k – 24 = 0

4k = 24

k = 6

Put the value of x in the equation.

3x² + 2kx — 3 = 0

–9 – 4k = 0

–4k = 9

Put x = 3/4

ax² + bx — 6 = 0

9a + 12b – 96 = 0 divide by 3

3a + 4b – 32 = 0

3a + 4b = 32 (1)

Put x = –2

ax² + bx — 6 = 0

a(–2)² + b(–2) – 6 = 0

4a – 2b – 6 = 0

4a – 2b = 6 (2)

Eliminate (1) and (2)

3a + 4b = 32

4a – 2b = 6 ×2

a = 4

Put a = 4 in equation (1).

3a + 4b = 32

3(4) + 4b = 32

12 + 4b = 32

4b = 32 – 12

4b = 20

b = 5

x² – (a + b)x + k = 0

Put x = a

a² – (a + b) a + k

a² – a² + ab + k = 0

ab + k = 0

k = –ab

(i) kx² – 5x + 6 = 0

Put x = 2

(ii) k² – 5(2) + 6 = 0

4k – 10 + 6 = 0

4k = 4

k = 1

(iii) 6x² + kx — = 0

Put x = – 2/√3

0

18 – 2√3 k – 4√6 = 0

18 – 4√6 = 2√3k

3√3 – 2√2 = k

(iv) ax² + bx – 10 = 0

Put

4a – 10b – 250 = 0

4a – 10b = 250 (1)

Put x = 3/5

ax² + bx – 10 = 0

25a + 15b – 90 = 0 (divide by 5)

5a + 3b = 18 (2)

Eliminate (1) and (2)

4a – 10b = 250 ×5

5a + 3b = 18 ×4

b = –19

Put b = –19 in (1)

4a – 10b = 250

4a – 10(–19) = 250

4a + 190 = 250

4a = 60

a = 15

2x– 2x – 3x + 3 = 0

2x (x– 1) – 3(x – 1) = 0

(2x – 3) (x – 1) = 0

2x – 3 = 0

x – 1 = 0

x =1

Therefore, the roots of the equation are , 1.

3x² – √6 x – √6 x + 2 = 0

3x² – √2√3x – √2√3x + 2 = 0

√3x(√3x – √2) – √2 (√3x – √2) = 0

√3x – √2 = 0 √3x – √2 = 0

Therefore, the roots of the equation are ,.

3x² — 15x + x — 5 = 0

3x (x – 5) + (x – 5) = 0

(3x + 1) (x – 5) = 0

3x + 1 = 0 x – 5 = 0

x=5

Therefore, the roots of the equation are ,.

Now, to find the roots by factorisation, we need to factorise 10 such that the sum is 10 and the product is 7√3 × √3 = 21

We can do that by 7 and 3.

So,

√3x² + 10x + 7√3 = 0

√3x² + 3x + 7x + 7√3 = 0

√3x(x + √3) + 7(x + √3) = 0

(√3x + 7)(x + √3) = 0

(√3x + 7) = 0

Or

x + √3 = 0

x = -√3

Hence, the solutions of the given quadratic equations are -√3 and .

√7 y² – 13y + 7y — 13 √7 = 0

y (√7 y – 13) + √7 (√7 y – 13) = 0

(√7 y – 13) (y + √7) = 0

√7 y – 13 = 0 y + √7 = 0

y = –√7

Rationalise

Therefore, the roots of the equation are ,.

4x² – {2(a² + b²) + 2(a² – b²)}x + (a² + b²) (a² – b²) = 0

4x² – 2(a² + b²)x + 2(a² – b²) x + (a² + b²) (a² – b²) = 0

2x {2x – (a² + b²)} – (a² – b²) {2x – (a² + b²)} = 0

{2x – (a² – b²)} {2x – (a² + b²)} = 0

2x – (a² – b²) = 0 2x – (a² + b²) = 0

2x = (a² – b²) 2x = (a² + b²)

Therefore, the roots of the equation are, .

b²x {a²x + 1} – 1 {a²x + 1} = 0

(b²x – 1) (a²x + 1) = 0

b²x – 1 = 0 a²x + 1 = 0

Therefore, the roots of the equation are.

(6x)² – 2 (6x)a + a² — b² = 0

Using Identity:

(x– y)² = x² + y² – 2xy

Here, (6x–a)² = (6x)² – 2 (6x) a + a²

(6x – a)² – b² = 0

Using identity:

x² – y² = (a+ b) (a – b)

(6x –a + b) (6x – a – b) =0

6x = a – b 6x = a + b

Therefore, the roots of the equation are,

2x (5ax – 3) + 3 (5ax – 3) = 0

2x+3 =0 5ax – 3 = 0

Therefore, roots of the equation are, .

12abx² — 9a² x— 8b²x — 6ab = 0

3ax (4bx – 3a) + 2b (4bx – 3a) = 0

3ax + 2b =0 4bx – 3a =0

Therefore, roots of the equation are, .

4x² — 2a²x+ 2b²x + a²b² = 0

2x (2x – a²) – b² (2x – a²) = 0

2x – b² = 0 2x – a² = 0

5x² – 6x – 2 = 0

Dividing by 5

We know

(a – b)² = a² – 2ab + b²

Here, a=x and –2ab =

–2xb = (∵ a =x)

–2b =

∴ Equation becomes

Add and subtract

Canceling squares both sides

Solving

So, and are the roots of the equation.

2x² – 5x + 3 = 0

Dividing by 2

Add a coefficient of to both sides

Dividing by 9

Add a coefficient of to both sides

x² – 9x = –18

Add the coefficient of to both sides

Add the coefficient of to both sides

Since root cannot be negative

Therefore, no real roots exist.

Add the coefficient of to both sides

x² – 6x = –4

Add the coefficient of to both sides

x² – 6x + (3)² = –4 + (3)²

(x – 3)² = –4 + 9

x – 3 = √5

x = ±√5 + 3

Divide by √5

Add the coefficient of to both sides

Divide by 2

Add the coefficient of to both sides

Since root cannot be negative

Therefore, it has no real roots.

x² + x = –3

Add the coefficient of to both sides

Since root cannot be negative

Therefore, it has no real roots.

Dividing by 5

Add the coefficient of to both sides

x = 2.6 + 2.4 x = –2.6 + 2.4

x = 5 x = – 0.2

Divide by 7

Add the coefficient of to both sides

Divide by 15

Add the coefficient of to both sides

Divide by 7

Add the coefficient of to both sides

d = b² – 4ac

d = (4)² – 4 (1) (3)

d = 16 – 12

d = 4

d = b² – 4ac

d = (5)² – 4 (4) (7)

d = 25 – 112

d = –87

d = b² – 4ac

d = (4)² – 4 (2) (5)

d = 16 – 40

d = –24

d = b² – 4ac

d = (5)² – 4 (3) (6)

d = 25 – 72

d = –47

d = b²– 4ac

d = (–2√2)² – 4 (√3) (–2√3)

d = 8 + 24

d = 32

TO CHECK REAL ROOTS, ‘d’ SHOULD BE GREATER THAN 0.

d = b² – 4ac

d = (8)²2 – 4 (7) (–1)

d = 64 – 28

d = 92

Yes, roots are real.

TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.

d = b²– 4ac

d = (3)² – 4(2)(4)

d = 9 – 32

d = – 23

No, roots aren’t real.

TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.

d = b² – 4ac

d = (–12)² – 4 (1) (–16)

d = 144 – 64

d = 208

Yes, roots are real .

TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.

d = b² – 4ac

d = (1)² – 4 (1) (–1)

d = 1 + 4

d = 5

Yes, roots are real.

TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.

d = b² – 4ac

d = (–10)² – 4 (1) (2)

d = 100 – 8

d = 92

Yes, roots are real.

REPEATED ROOTS MEAN d = 0.

d = b² – 4ac

d = (–12)² – 4(9)(4)

d = 144 – 144

d = 0

Yes, roots are repeated.

REPEATED ROOTS MEAN d = 0.

d = b² – 4ac

d = (–6)² – 4(1)(6)

d = 36 – 24

d = 12

∴ roots are not repeated.

REPEATED ROOTS MEAN d = 0.

d = b² – 4ac

d = (4)² – 4 (9) (6)

d = 16 – 216

d = –200

∴ roots are not repeated.

REPEATED ROOTS MEAN d = 0 .

d = b² – 4ac

d = (–40)² – 4 (16) (25)

d = 1600 – 1600

d = 0

∴ roots are repeated.

REPEATED ROOTS MEAN d = 0 .

d = b² – 4ac

d = (6)² – 4(1)(9)

d = 36 – 36

d = 0

∴ roots are repeated.

d = b² – 4ac

d = (7)² – 4 (4) (2)

d = 49 – 32

d = 17

Since, d>0, roots are unique and real.

d = b² – 4ac

d = (10)² – 4 (1) (39)

d = 100 – 156

d = –56

Since, d < 0, no real roots exists.

d = b² – 4ac

d = (–4)² – 4 (2) (3)

d = 16 – 24

d = –8

Since, d < 0, no real roots exist for the given equation.

d = b² – 4ac

d = 0

Since, d = 0, roots are real and equal for the given equation.

d = b² – 4ac

d = (–6)² – 4 (2) (3)

d = 36 – 24

d = 12

∴ Roots are real and unique.

d = b² – 4ac

d = (–5)² – 4 (2) (–3)

d = 25 + 24

d = 49

∴ Roots are real and unique.

Since roots are equal

∴ d=0 ….(1)

4x² — 2 (k + 1) x + (k + 4) = 0

d = b² – 4ac

d = (–2 (k + 1)² – 4 (4) (k + 4)

d = (– 2k – 2)² – 16k – 64

d = 4k² + 4 + 8k – 16k – 64

(∵ (a – b)² = a² + b² – 2ab)

d = 4k² – 8k – 60

From (1), d = 0

∴ Equation will be:

4k² – 8k – 60 = 0

Dividing by 4

k² – 2k – 15 = 0

4k² – 5k + 3k – 15 = 0

k (k – 5) + 3(k – 5) = 0

(k – 5) (k + 3) = 0

K – 5 = 0 k + 3 = 0

K = 5 k = –3

Since roots are equal

∴ d=0 ….(1)

(k + 1)x² — 2 (k – 1) x + 1 = 0

d = b² – 4ac

d = (–2(k–1))²– 4(k+1)(1)

d = (–2k+2)² – 4k – 4

d=4k² + 4 – 8k – 4k – 4

(∵ (a + b)² = a² + b² + 2ab)

d = 4k² – 12k

From (1), d = 0

∴ Equation will be:

0 = 4k² – 12k

4k² = 12k

k² = 3k

k² – 3k = 0

k(k – 3) = 0

k = 0 or k – 3 = 0

k = 3

∴ Values of k are 0, 3.

Since roots are equal

∴ d=0 (1)

9x² + 8kx + 16 = 0

d = b² – 4ac

d = (8k)² – 4 (9) (16)

From (1), d = 0

∴ Equation will be:

0 = 64k² – 576

k² = 9

k = ±√9

k = 3, –3

∴ values of k are –3, 3 .

Since roots are equal

∴ d=0 (1)

(k + 4)x² + (k + 1)x + 1 = 0

d=b²–4ac

d = (k – 1)²– 4 (k + 4) (1)

d = (–2k + 2)² – 4k – 4

d = k² + 1+ 2k – 4k – 16

From (1), d = 0

∴ Equation will be:

0 = k² + 1 + 2k – 4k – 16

k² – 2k – 15 = 0

k² – 5k + 3k – 15 = 0

k(k – 5) + 3 (k – 5) = 0

(k – 5) (k + 3) = 0

K – 5 = 0 k + 3 = 0

k = 5 k = –3

∴ Values of k are –3, 5.

Since roots are equal

∴ d=0 (1)

k²x² — 2(2k — 1)x + 4 = 0

d = b² – 4ac

d = (–2(k–1))² – 4 (k²) (4)

d = (–2k+2)² – 4k – 4

d = (–4k+2)² – 16k²

(∵ (a – b)² = a² + b² – 2ab)

d = 16k² – 16k + 4 – 16k²

d = –16k + 4

From (1), d = 0

∴ Equation will be:

0 = –16k + 4

16k = 4

∴ Values of k are is .

Since roots are equal

∴ d=0 (1)

(a — b)x² + (b — c) x + (c — a) = 0

d = b² – 4ac

d = (b–c)² – 4 (a–b) (c–a)

d = b² + c² – 2bc –4 [a (c – a) – b (c – a)]

d = b² + c² – 2bc – 4 [ac – a² – bc + ba]

From (1), d = 0

∴ Equation will be:

0 = b² + c² – 2bc – 4ac + 4a² + 4bc – 4ba

b² + c² – (2a)² 2bc + 2c (–2a) + 2(–2a)b = 0

(b + c – 2a)² = 0

(b + c – 2a) = 0

b + c = 2a

Hence proved.

2x² + 2px — 15 = 0

Put x = –5

2(–5)² + 2p(–5) — 15 = 0

50 – 10p – 15 = 0

35 = 10p

p = 3.5

Equation p (x² + x) + k = 0 has equal roots i.e. d = 0

p (x² + x) + k = 0

p = 3.5

3.5 (x² + x) + k = 0

3.5x² + 3.5x + k = 0

d = b² – 4ac

d = (3.5)² – 4(3.5)(k)

d = 12.25 – 14k

Putting d = 0

∴ Equation will be:

0 = 12.25 – 14k

14k = 12.25

k = 0.875

Roots are equal

∴ d = 0

d = b² – 4ac

d = (–10)² – 4 (2) (k)

d = 100 – 8k

Put d = 0

0 = 100 – 8k

8k = 100

Roots are equal

∴ d = 0

d = b² – 4ac

d = (–6)² – 4 (–2) (k)

d = 36 + 8k

Put d = 0

0 = 36 + 8k

–8k = 36

Roots are equal

∴ d = 0

d = b² – 4ac

d = (4)² – 4 (1) (k)

d = 16 – 4k

Put d = 0

0 = 16 – 4k

4k = 16

k = 4

Roots are equal

∴ d = 0

d = b² – 4ac

d = (–2√5)² – 4 (k) (4)

d = 20 – 16k

Put d = 0

0 = 20 – 16k

16k = 20

Roots are equal

∴ d = 0

d = b² – 4ac

d = (4k)² – 4 (k² – k + 2) (1)

d = 16k² – 4k² + 4k – 8

Put d = 0

0 = 16k² – 4k² + 4k – 8

12k² + 4k² + 4k – 8 = 0 (divide by 4)

3k² + k – 2 = 0

3k² + 3k – 2k – 2 = 0

3k (k + 1) – 2k (k + 1) = 0

(3k–2k)(k+1) = 0

(3k–2k)=0 or (k+1) = 0

k = 0 k = –1

x²(a² + b²) + 2x(ac + bd) + (c² + d²)= 0

d = b² – 4ac

d = (2ac + 2bd)² – 4 (a² + b²) (c² + d²)

d = 4a²c² + 4b²d² + 8abcd – 4 [a² (c² + d²) + b² (c²+d²)]

d = 4a²c² + 4b²d² + 8abcd – 4 [a²c² + a²d² + b²c² + b²d²]

d = 4a²c² + 4b²d² + 8abcd – 4a²c² – 4a²d² – 4b²c² – 4b² d²

d = 8abcd – 4a²d² – 4b²c²

d = 8abcd – 4(a²d² + b² c²)

d = –4 (a² d² + b²c² – 2abcd)

d = –4 [(ad + bc)²]

For ad ≠ bc

d= –4 × [value of (ad + bc)²]

∴ d is always negative

So, d < 0

The given equation has no real roots.

Let the first number be ‘X’, so the other number will be ’(12–X)’.

∵ X (12 – X) = 32

12X – X² = 32

X² – 12X + 32 = 0

On factorising further,

X² – 4X – 8X + 32 = 0

X(X – 4) – 8(X – 4) = 0

(X – 4)(X – 8) = 0

So, X = 4 or 8

∴ The numbers are 4 and 8.

Let the first number be ‘X’, so the other number will be ’(X+3)’.

∵ X(X + 3) = 504

X² + 3X– 504=0

On applying Sreedhracharya formula

X = –24 or 21.

∴ if the first number is – 24 , then the other number is – 21.

∴ if the first number is 21 , then the other number is 24.

∴ The numbers are not (21,24) & (– 21, – 24).

Let the first number be ‘X’, so the other number will be ’(X+1)’.

∵ X²+(X+1)² = 365

X² + X² + 1 + 2X – 365 = 0

2X² + 2X – 364 = 0

X² + X – 182 = 0

On applying Sreedhracharya formula

∴ X = –14 or X = 13.

∴ The numbers are 13 & 14.

Let the first number be ‘X’, so the other number will be ’(X+4)’.

On simplifying further,

X² + 4X – 21 = 0

On applying Sreedhracharya formula

∴ X = –7 or X = 3.

∴ The numbers are –7 & –3 or 3 & 7.

Let the first number be ‘X’, so the other number will be ’(18–X)’.

On simplifying further,

–X² + 18X – 72 = 0

X² – 18X + 72 = 0

On applying Sreedhracharya formula

X = 6 or X = 12

∴ The numbers are 6 & 12.

Let the first number be ‘X’, so the other numbers will be ’(X+1)’ & ‘(X+2).

∵X² + (X + 1)² + (X + 2)² = 50

X² + X² + 1 + 2X + X² + 4 + 4X = 50

On simplifying further,

3X² + 6X– 45 = 0

X² + 2X – 15 = 0

On applying Sreedhracharya formula

X = – 5 or 3

∴ X = 3 (Only Positive values)

∴ X + 1 = 4

∴ X +2 = 5

∴ The numbers are 3, 4 & 5.

Let the first number be ‘X’, so the other numbers will be ’(X+1)’ & ‘(X+2).

∵ X² + (X + 1)(X + 2) = 154

X² + X² + 2X + X + 2 = 154

On simplifying further,

2X² + 3X – 152 = 0

On applying Sreedhracharya formula

X = – 9.5 or 8

∴ X = 8 (Only whole values)

∴ X +1 = 9

∴ X + 2 = 10

∴ The numbers are 8, 9 & 10.

Let the units digit be ‘Y’, so the tens digit will be X.

∴ the number is 10X + Y

X . Y =14 –––––– (i)

On simplifying further,

9X – 9Y + 45 = 0

X – Y + 5 = 0

Putting the value of from equation –––––– (i)

14 – Y² + 5Y = 0

Y² – 5Y – 14 = 0

On applying Sreedhracharya formula

Y = – 2 or 7

∴ Y = 7 (Only Positive values)

∴ X = 2

∴ The number is 27 {2 (10) + 7}.

Let the larger number be ‘X’, so the smaller number will be ‘Y’.

∴ X² – Y² = 45 ––––– (i)

∵4X = Y² –––––– (ii)

On simplifying further,

Putting the value of Y² = 4X from equation –––––– (ii)

X² – 4X = 45

X² – 4X – 45 = 0

On applying Sreedhracharya formula

X = – 5 or 9

∴ X = 9 (Only natural number, as given in the question)

∴ Y=√(4*9)

∴ Y = 6

∴ The numbers are 9 & 6.

Let the first number be ‘X’, so the other number will be ’(X+5)’.

On simplifying further,

X² + 5X – 50 = 0

On applying Sreedhracharya formula

∴ X = –10 or X = 5.

Then other numbers will be (X+5) = {–10+5} & {5+5} i.e., –5 or 10.

∴ The numbers are –10 & –5 or 5 & 10.

Let the number be ‘X’, so the reciprocal will be ’’.

On simplifying further,

3 + 3X² = 10X

3X² – 10X + 3 = 0

On applying Sreedhracharya formula

∴ X = 3 or

Then other numbers will be & 3

∴ The numbers are 3 or

Let the first number be ‘X’, so the other number will be ’(12–X)’.

∵ X² + (12 – X)² = 74

X² + 144 + X² – 24X = 74

On simplifying further,

2X² – 24X + 70 = 0

X² – 12X + 35 = 0

On applying Sreedhracharya formula

∴ X = 5 or X = 7.

Then other numbers will be (12–X) = {12–5} & {12–7} i.e., 7 or 5.

∴ The number 12 is divided into two parts namely 5 & 7.

Let the first number be ‘X’, so the other number will be ’(X+1)’.

∵ X² + (X + 1)² = 421

X² + X² + 1 + 2X = 421

On simplifying further,

2X²+ 2X – 420 = 0

X² + X – 210 = 0

On applying Sreedhracharya formula

∴ X = –15 or X = 14.

∴ X = 14 (Only natural number, as given in the question)

∴ Then other numbers will be 14 & 15.

Let the units digit be ‘Y’, so the tens digit will be X.

∴ the number is 10X + Y

∵ X . Y =18 –––––– (i)

10X+Y–63=10Y+X

On simplifying further,

9X – 9Y – 63 = 0

X – Y – 7 = 0

Putting the value of from equation –––––– (i)

18 – Y² – 7Y = 0

Y² + 7Y – 18 = 0

On applying Sreedhracharya formula

Y = – 9 or 2

Y = 2 (Only Positive values)

∴ X = 9

∴ The number is 92 {9(10)+2}.

Let the units digit be ‘Y’, so the tens digit will be X.

the number is 10X + Y

10X + Y = 5 (X + Y)

On simplifying further,

10X – 5X – 5Y + Y = 0

5X – 4Y = 0 –––––– (i)

10X + Y = 2XY + 5

10X– 2XY + Y – 5 = 0 –––––– (ii)

Putting the value of from equation –––––– (i)

40Y – 8Y² + 5Y – 25 = 0

8Y² – 45Y + 25 = 0

On applying Sreedhracharya formula

Y = or 5

∴ Y = 5

∴ from equation (i)

∴ X = 4

∴ The number is 45 {4(10)+5}.

Let the numerator be ‘X’, so the denominator will be ‘(2X+1)’.

the fraction is

On simplifying further,

21(5X² + 4X + 1) = 58 (2X² + X)

105X² + 84X + 21 = 116X² + 58X

11X² – 26X – 21 = 0–––––– (i)

On applying Sreedhracharya formula

X= or 3

∴ X = 3 (Only positive values)

∴ Numerator is = 3 & denominator is (2X+1) = 7

∴ The fraction is .

Given: Numerator of a fraction is one more than its denominator.

To Find: The fraction

Assumption: Let the denominator be x.

Numerator = x + 1

Therefore, the fraction

From the second case, we get,

Taking L.C.M we get,

Cross-multiplying we get,

30(2x + 1) = 11(x² + x)

60x + 30 = 11x² + 11x

11x² + 11x – 60x – 30 = 0

11x² – 49x – 30 = 0

Now, we need to factorise such that, on multiplication we get 330 and on substraction we get 49.

Therefore, 55 and 6 can be the factors.

So, equation becomes,

11x² – (55x – 6x) – 30 = 0

11x² – 55x + 6x – 30 = 0

11x(x – 5) + 6(x – 5) = 0

(11x + 6)(x – 5) = 0

So, 11x + 5 = 0 or x – 5 = 0

So the possible fractions are:

Or

Let the denominator be ‘X’, so the numerator will be ‘(X+1)’.

the fraction is

On simplifying further,

30(2X² + 2X + 1) = 61 (X² + X)

60X² + 60X + 30 = 61X²+ 61X

X² + X – 30 = 0–––––– (i)

On applying Sreedhracharya formula

X= – 6 or 5

∴ X = 5 (Only positive values)

∴ Denominator is = 5 & numerator is (X+1) = 6

∴ The fraction is .

Let the denominator be ‘X’, so the numerator will be ‘(X+3)’.

the fraction is

On simplifying further,

28(6X + 9) = 33(X² + 3X)

168X + 252 = 33X² + 99X

33X² – 69X – 252 = 0

11X² – 13X – 84 = 0 –––––– (i)

On applying Sreedhracharya formula

it does not have any real values.

Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.

the original fraction is

the new fraction is

On simplifying further,

24(3) = X² + 7X + 12

X² + 7X – 60 = 0 –––––– (i)

On applying Sreedhracharya formula

∴ X = –12 or 5.

∴ the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be , as taking (–12) will form a fraction i.e., (not satisfying the conditions) .

Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.

the original fraction is

the new fraction is

On simplifying further,

88(9) = 9(X² + 9X + 18)

X² + 9X – 70 = 0 –––––– (i)

On applying Sreedhracharya formula

X = –14 or 5.

the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be , as taking (–14) will form a fraction i.e., (not satisfying the conditions) .

Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.

the original fraction is

the new fraction is

On simplifying further,

15(2X² + 10X + 6) = 19(X² + 8X + 15)

30X² + 150X + 90 = 19X² + 152X + 285

11X² – 2X – 195 = 0 –––––– (i)

On applying Sreedhracharya formula

∴ it does not have real values.

Given: Numerator of a fraction is 2 less than the denominator

To find: The fraction

Assumption: Let the denominator be x

Numerator = x – 2

Therefore, the fraction

If one is added to the numerator and denominator, fraction becomes

Sum of the fractions

Sum of the fractions

Therefore,

Taking L.C.M we get,

Cross-multiplying we get,

30x² – 30x – 30 = 19x² + 19x

30x² – 19x² – 30x – 19x – 30 = 0

11x² – 49x – 30 = 0

Now we need to factorise such that, on multiplication we get 330 and on substraction we get 49.

So, equation becomes,

11x² – (55x – 6x) – 30 = 0

11x² – 55x + 6x – 30 = 0

11x(x – 5) + 6(x – 5) = 0

(11x + 6)(x – 5) = 0

So, 11x + 5 = 0 or x – 5 = 0

Putting these values in fraction

Hence, the possible fractions are,

Let the shortest side(AC) be ‘(X)’cms, so the hypotenuse (BC) will be ‘(2X+6)’ cms, as demonstrated in the figure drawn below:

∴ AB = 2X + 4 cms

∵ (BC)² = (AC)² + (AB)²

∴(2X + 6)² = X² +(2X + 4)²

On simplifying further,

4X² + 24X + 36 = X² + 4X² + 16X + 16

Using the identity of a² + b² + 2ab = (a + b)²

X² – 8X – 20 = 0 –––––– (i)

On applying Sreedhracharya formula

∴ X = –2 or 10.

∴ the shortest side (AC) is X = 10 cms (Only positive values), hypotenuse (2X+6) i.e., (BC) is 26 cms and the other side (AB) is (2X+4) i.e., 24 cms .

Let the sides of the squares be ‘(X)’cms and ‘(Y) cms.

∴ X² + Y² = 640 –––––(i)

∵ Area = (Side)²

∵Perimeter = 4 (Side)

∴4X – 4Y = 64

On simplifying further,

X – Y = 16 ––––––(ii)

Squaring the above mentioned equation, i.e., equation (ii)

X² + Y² – 2XY = 256

Using the identity of a² + b² – 2ab = (a – b)²

Putting the value of equation (i) in equation (ii)

∴ 640 – 2XY = 256

2XY = 384

XY = 192

Putting the value of in equation (i)

(Y+16)² + Y² = 640

Y² + 256 + 32Y + Y² = 640

2Y² + 32Y – 384 = 0

Y² + 16Y – 192 = 0

On applying Sreedhracharya formula

∴ X = – 12 or 24.

∴ the side is X = 24 cms (Only positive values), other square’s side is Y=X–16 i.e., 8 cms.

Let the shortest side(AC) be ‘(X)’cms, and the longer side (AB) be ‘(Y)’ cms, as demonstrated in the figure drawn below:

∴ BC = 3√10 cms

∵ (BC)² = (AC)² + (AB)²

∴ (3√10)² = X² + (Y)²

On simplifying further,

(Y)² = 90 – X²

X² + Y² – 90 = 0 –––––– (i)

As per the question,

New smaller side = ‘(3X)’ cms

New longer side = ‘(2Y)’ cms

∴ BC = 9√5 cms

∵ (BC)² = (AC)² + (AB)²

∴(9√5)² = (3X)² + (2Y)²

On simplifying further,

4Y² + 9X² = 405

4X² + 9Y² – 405 = 0 –––––– (ii)

Putting the value of X², from equation (i) in equation (ii)

X² = 90 – Y²

On simplifying further,

(360–4Y)² + 9Y² – 405 = 0

5Y² = 45

Y = ± 3 cms(Only positive values),

∴ X = √(90–9).

∴ X = 9 cms

the shortest side (AC) is X = 9 cms hypotenuse i.e., (BC) is cms and the other side (AB) is 3 cms .

Ok now we know that area of square = side²

So no. of students in the line if side was x (assumption)

No. of students = x²+24 (as there were 24 exrltra students)

No. Of students after increasing 1 student in square = (x+1)²–25 (as there were 25 less students)

So x² + 24 = (x + 1)² – 25

x²+ 24 = x² + 1 + 2x – 25

x² – x² + 24 + 25 – 1 = 2x

48 = 2x

So x = 24

So no. Of students = x² + 24 = 24² + 24 = 576 + 24 = 600

the no. of students = 600.

Let the length of base = P cm.

As base exceeds the base by 7cm ,

then , length of altitude = (P + 7)cm

now, area of triangle =

given, area of triangle = 30 cm²

so,

⇒ 30 × 2 = P² + 7P

⇒ 60 = P² + 7P

⇒ P² + 7P – 60 = 0

⇒ P² + 12P – 5P – 60 = 0

⇒ P(P + 12) – 5(P + 12) = 0

⇒ (P + 12)(P –5) = 0

⇒ P = 5 , –12

but length can't be negative so, P ≠ –12

hence, P = 5 cm e.g., base = 5cm

Let breadth be X cm
length = 2X cms
area=800
2X (X)=800
X(X)=400

X² = 400
X =20
Yes it is possible to design a rectangular mangrove having breadth =20 cm and length=40 cm.

Let the length be

then breadth= l – 3

area of rectangle= l(l – 3) = l² – 3l

area of triangle=

given

area of rectangle is 4sq mt more than triangle

so

area of rectangle – 4= area of triangle

l² – 3l – 4 = 6l – 18

l² – 9l + 14 = 0

l² – 7l – 2l + 14 = 0

l(l – 7) – 2(l – 7) = 0

(l – 7)(l – 2) = 0

Yes, it is possible, to design a rectangular park.

L = 7 and 2

L = 2 is neglected as when length is 2 the breadth will be negative which is not possible...

so length= 7m

breadth= =7–3 = 4m

Yes it is possible to design a rectangular park having length & breadth 7 mts & 4 mts respectively.

Let P be the position of the pole and A and B be the opposite fixed gates.

PA = ‘a’ mts

PB = ‘b’ mts

PA – PB = 7 m

⇒ a – b = 7

⇒ a = 7 + b .........(1)

In Δ PAB,

AB² = AP² + BP²

⇒ (17) = (a)² + (b)²

⇒ a² + b² = 289

⇒ Putting the value of a = 7 + b in the above,

(7 + b)² + b² = 289

⇒ 49 + 14b + 2b² = 289

⇒ 2b² + 14b + 49 – 289 = 0

⇒ 2b² + 14b – 240 = 0

Dividing the above by 2, we get.

⇒ b² + 7b – 120 = 0

⇒ b² + 15b – 8b – 120 = 0

⇒ b(b + 15) – 8(b + 15) = 0

⇒ (b – 8) (b + 15) = 0

⇒ b = 8 or b = –15

Since this value cannot be negative, so b = 8 is the correct value.

Yes it is possible to erect a pole.

Putting b = 8 in (1), we get.

a = 7 + 8

a = 15 m

Hence PA = 15 m and PB = 8 m

So, the distance from the gate A to pole is 15 m and from gate B to the pole is 8 m.

Let the mother age be ‘X’ years, then her daughter’s age will be ‘(20–X)’ years.

As per the question,

(X – 4)(20 – X – 4) = 48

(X – 4)(16 – X) = 48

16X – 64 – X² + 4X = 48

–X² + 20X – 112 = 0

X² – 20X + 112 = 0

On applying Sreedhracharya formula

it does not have real values, then it is not possible for the above situation to happen.

Let the speed of the train and the time taken to cover the same be ‘X’ km/hr and ‘Y’ hrs respectively.

As per the question,

XY = 90 –––––– (i)

(X + 15)(Y – 0.5) = 90 ––––– (ii)

LHS = RHS

Equating the LHS of both equations, and simplyifying it further

XY = XY – 0.5X + 15Y – 7.5

0.5X – 15Y + 7.5 = 0 ––––––– (iii)

Multiplying the above equation by 10,

5X – 150Y + 75 = 0

Simplyfying it further,

X – 30Y + 15 = 0

Putting the value of Y, in equation (iii)

X² + 15X – 2700 = 0

On applying Sreedhracharya formula

X = – 60 or X = 45

the original speed of train is 45 km/hr as speed can’t be negative.

Let the usual speed be x km /hr.

Actual speed = (x + 250) km/hr.

Time taken at actual speed = () hr.

Difference between the two times taken = hr.

Speed at that time = (x + 250) km/hr

Then, According to the question,

3000{250} = x² + 250

0 = x² + 250x – 750000

0 = x² + (1000–750)x – 750000

0 = x² + 1000x – 750x – 750000

0 = x(x + 1000) – 750(x + 1000)

0 = (x + 1000) (x – 750)

x = –1000 or x = 750

Usual speed = x = 750 km /hr

⇒ x = 750 【 speed cannot be negative】

Hence , the usual speed of the aeroplane was 750 km / hr.

Let the speed of the boat be ‘X’ km/ hr, time taken for upstream and downstream be T1 hrs & T2 hrs respectively.

––––––– (i) (Downstream)

––––––– (ii) (Upstream)

Adding equation equation (i) & equation (ii),

2025 – 9X² = 1800

9X² = 225

X² = 25

X = ±5

∵ speed can’t be negative.

∴ The speed of the stream be 5 Km/hr

Let the average speed of the passenger train be 'x' km/hr and the average speed of the express train be (x + 11) km/hr

Distance between Mysore and Bangalore = 132 km

It is given that the time taken by the express train to cover the distance of 132 km is 1 hour less than the passenger train to cover the same distance.

So, time taken by passenger train = hr

The time taken by the express train = hr

Now, according to the question

After taking L.C.M. of and then solving it we get .

Now,

By cross multiplying, we get

132x = x² + 132x + 11x + 1452

x² + 11x – 1452 = 0

x² + 44x – 33x – 1452 = 0

x(x + 44) – 33(x + 44) = 0

(x + 33) (x + 44) = 0

x – 44 or x = 33

As the speed cannot be in negative therefore, x = 33 or the speed of the passenger train = 33 km/hr and the speed of express train is 33 + 11 = 44 km/hr.

Let the present age of Rehman be x years.

So, 3 years age his age was = (x – 3) years

The reciprocal =

And, after 5 years the age will be = (x + 5) years

The reciprocal =

So, according to the question

Taking L. C. M. of (x – 3) and (x + 5)

6x + 6 = x² + 2x – 15

x² + 2x – 6x – 15 – 6 = 0

x² – 4x – 21 = 0

x² – 7x + 3x – 21 = 0

x(x – 7) + 3(x – 7) = 0

(x – 7)(x + 3) = 0

x = 7 and x = –3

x = – 3 is not possible because age cannot be negative.

So, x = 7

Therefore present age of Rehman is 7 years.

Let the father and his son’s age be ‘X’ yrs and ‘Y’ yrs respectively.

X + Y = 35
X × Y = 150

150 + Y² = 35Y
Y² – 35Y + 150 = 0
(Y–5) (Y–30)=0
Y=5
the son's age (Y) = 5 yrs and father's age (X) = 30 yrs.

Let the father and his son’s age be ‘X’ yrs and ’(24–X)’ yrs respectively.

As per the question,

24X – X² – 96 + 4X = 36

X² – 28X + 132 = 0

X² – 22X – 6X + 132 = 0

X(X – 22) – 6(X – 22) = 0

(X – 6)(X – 22) = 0

∴X = 22 & boy’age is 2 years.

∴X = 6 & boy’age is 18 years. (practically not possible)

The age of father and his son are 22 years & 2 years respectively.

Let the sister and her sister’s age be ‘X’ yrs and ’(X+5)’ yrs respectively.

As per the question,

X(X + 5) = 104

X² + 5X – 104 = 0

X² + 13X – 8X + 132 = 0

X(X + 13) – 8(X + 13) = 0

(X + 13)(X – 8) = 0

∴X = 8 or X = –13

the ages of sisters are 8 years & 13 years respectively, as the age can’t be negative.

Let seven years, age of Swati was x years and age of Varun was years.

Present age of Swati = (x + 7) years

Present age of Varun = (5x² + 7) years

Given, after 3 years swati's age will be th of Varun's age.

Age of Swati after 3 years = (x + 7 + 3) years = (x + 10) years

Age of Varun after 3 years = (5x² + 7 + 3) years = (5x² + 10) years

Given, age of Swati after 3 years = Two–fifths age of varun after 3 years

(x + 10) = 2(x² + 2)

x + 10 = 2x² + 4

2x² – x + 4 – 10 = 0

2x² – x – 6 = 0

2x² – 4x + 3x – 6 = 0

2x(x – 2) + 3(x – 2) = 0

x = 2 [ Age can't be negative ]

Therefore, the present age of Swati = (2 + 7) = 9 year and the present age of Varun = 5(2)²+7=27 years.

Let the marks scored in maths be ‘X’.

Marks in English is ‘(40–X)’.

As, per the question,

If he got 3 marks in maths & 4 marks less in English,

Marks in Maths =X+3

Marks in English = 40–X–4 = 36–X

Product = 360

(36 – X)(X + 3) = 360

(36X + 108 – X² – 3X) = 360

(33X + 108 – X²) = 360

X² – 33X + 360 – 108 = 0

X² – 33X + 252 = 0

X² – 21X – 12X + 252 = 0

X(X – 21) – 12(X – 21) = 0

(X– 12)(X – 21) = 0

X = 12 or 21

∴ If marks in Maths = 12 then marks in English = 40 – 12 = 28

If marks in Maths = 21 then marks in English = 40 – 21 = 19

Let the marks in maths be ‘X’ and English be ‘Y’.

X + Y = 45 equation 1

X = 45 – Y

(X + 1)(Y – 1) = 500

(45 – Y – 1)(Y – 1) = 500

Y² – 43Y + 456 = 0

By solving this quadratic equation

Y² – 24Y – 19Y + 456 = 0

Y(Y – 24) – 19(Y – 24) = 0

we get two values of Y

Y1 = 24

Y2 = 19

substitute both this values in equation 1

X1 + 24 = 45

X1 = 45 – 24

= 21

X2 + 19 = 45

X2 = 45 – 19

= 26

the marks in maths is 21, then marks in English is 24 and if the marks in maths is 26, then marks in English is 19.

Let x be the no. of person and y is amount taken by each person.

……(1)

when 15 more person appear then

……(2)

On solving both (1) and (2) equation

subtracting (2) from (1)

3250 = x² + 15x

x² + 15x – 3250 = 0

On applying Sreedhracharya formula

X = – 65 or X = 50

The persons are 50 (As positive values are only considered).

Let x be the no. of students and y is the number of apples taken by each person.

––––––(1)

when 10 more students appear then,

–––––(2)

On solving both (1) and (2) equation

subtracting (2) from (1)

3000 = x² + 10x

x² + 10x – 3000 = 0

On applying Sreedhracharya formula

X = – 60 or X = 50

The number of students are 50. (as only positive values are considered).

Let x be the no. of books and y is the cost of each book.

––––––(1)

when 10 more students appear then,

––––(2)

On solving both (1) and (2) equation

subtracting (2) from (1)

600 = x² + 10x

x² + 10x – 600 = 0

On applying Sreedhracharya formula

X = – 30 or X = 20

The number of books, he purchased are 20. (as only positive values are considered).

Let the total number of camels be x.

Number of camels in forest = .

Number of camels gone to mountains = 2√x

Remaining camels on bank of the river = 15

Total camels =

4x = x + 8√x + 60

3x – 60 = 8√x

Squaring both the sides,

9x² + 3600 – 360x = 64x

9x² – 424x + 3600 = 0

On simplifying further,

9x² – 100x – 324x + 3600 = 0

x(9x – 100) – 36(9x – 100) = 0

(x – 36)(9x – 100) = 0

x = or X = 36

The number of camels is 36. (As whole values are only considered)

Amount for the booking of hotel = Rs. 1200

Let there be x tourists

When all the tourist paid money, each share = Rs.

When 3 members failed to pat, its Rs.

So, according to the question:

3600 = 20 × x(x – 3)

3600 = 20x²– 60x

20x² – 60x – 3600 = 0

On simplifying further,

x² – 3x – 180 = 0

x² – 15x + 12x – 180 =0

x(x – 15) + 12(x – 15) = 0

Therefore, x = –12 or x = 15

The number of camels is 15. (As positive values are only considered)

Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+5)’ minutes to fill it.

Applying the concept of Unitary Method,

In one minute, both pipes will fill the part of cistern as below:

12X + 30 = X² + 5X

X² – 7X – 30 = 0

On factorising the same.

X² – 10X + 3X – 30 = 0

X(X – 10) + 3(X – 10) = 0

(X – 10)(X + 3) = 0

∴ X = – 3 or X = 10.

Then the first pipe will fill the cistern in ‘X’ minutes i.e., 10 minutes and the second pipe will fill the cistern in ‘(X+5)’ minutes i.e., 15 minutes.

Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+1)’ minutes to fill it.

Applying the concept of Unitary Method,

In one minute, both pipes will fill the part of cistern as below:

60X + 30 = 11X² + 11X

11X² – 49X – 30 = 0

On applying Sreedhracharya formula

X = – or X = 5

The time required to fill the cistern is 5 & 6 minutes respectively. (as only positive values are considered).

Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+3)’ minutes to fill it.

Applying the concept of Unitary Method,

In one minute, both pipes will fill the part of cistern as below:

80X + 120 = 13X² + 39X

13X² – 41X – 120 = 0

On applying Sreedhracharya formula

X = or X = 5

The time required to fill the cistern is 5 & 8 minutes respectively. (as only positive values are considered).