Quadratic Equations

(x + 1)² = 2(x - 3)

LHS = (x + 1)² = x² + 1 + 2x

RHS = 2(x – 3) = 2x – 6

∴ x² + 1 + 2x = 2x - 6

⇒ x² + 1 + 2x – 2x + 6 = 0

⇒ x² + 7 = 0 which is of the form ax² + bx + c = 0

∴ it is a quadratic equation.

x² - 2x = ( - 2)(3 - x)

LHS = x² – 2x

RHS = ( - 2)(3 - x) = - 6 + 2x

∴ x² - 2x = - 6 + 2x

Or x² - 2x + 6 – 2x = 0

⇒ x² + 6 which is of the form ax² + bx + c = 0

∴ it is a quadratic equation.

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

LHS = (x - 2)(x + 1) = x² + x - 2x - 2 = x² - x - 2

RHS = (x - 1) (x + 3) = x² + 3x - x - 3 = x² + 2x - 3

Equating LHS and RHS

x² - x - 2 = x² + 2x - 3

⇒ - 3x + 1 = 0 which is not of the form ax² + bx + c = 0

Hence it is not a quadratic equation.

(x - 3)(2x + 1) = x(x + 5)

Here LHS = (x - 3)(2x + 1) = 2x² - 6x + x - 3 = 2x² - 5x - 3

RHS = x(x + 5) = x² + 5x

Equating LHS and RHS

2x² - 5x - 3 = x² + 5x

x² - 10x - 3 = 0 which is of the form ax² + bx + c = 0

∴ it is a quadratic equation.

(2x - 1) (x - 3) = (x + 5)(x - 1)

Here LHS = (2x – 1)(x - 3) = 2x² –x - 6x + 3 = 2x² - 7x + 3

RHS = (x + 5)(x - 1) = x² + 5x – x - 5 = x² + 4x - 5

Equating LHS and RHS

2x² - 7x + 3 = x² + 4x - 5

x² – 11x + 8 = 0

which is of the form ax² + bx + c = 0

∴ it is a quadratic equation.

x² + 3x + 1 = (x - 2)²

Here RHS = (x - 2)²

⇒ x² + 4 - 4x

Equating LHS and RHS

x² + 3x + 1 = x² + 4 - 4x

7x - 3 = 0

Which is not of the form ax² + bx + c = 0

∴ it is not a quadratic equation.

(x + 2)³ = 2x(x² - 1)

LHS = (x + 2)³ = x³ + 2³ + 3 × x × 2(x + 2)

(∵(a + b)³ = a³ + b³ + 3ab(a + b))

= x³ + 8 + 6x² + 12x

RHS = 2x³ - 2x

Equating LHS and RHS

x³ + 8 + 6x² + 12x = 2x³ - 2x

- x³ + 6x² + 14x + 8 = 0

which is not of the form ax² + bx + c = 0

∴ it is not a quadratic equation.

x³ - 4x² - x + 1 = (x - 2)³

RHS = (x - 2)³ = x³ - 2³ - 3 × x × 2(x - 2)

(∵(a + b)³ = a³ + b³ + 3ab(a + b))

= x³ - 8 - 6x² + 12x

Equating LHS and RHS

x³ - 4x² –x + 1 = x³ - 8 - 6x² + 12x

2x² - 13x + 9 = 0

which is of the form ax² + bx + c = 0

∴ it is a quadratic equation.

Let length = L and breath = B

Area of rectangle (A) = L × B

According to the given conditions

L = 2B + 1

∴ A = (2B + 1) B

⇒ 528 = 2b² + b

⇒ 2b² + b - 528 = 0

Which is a quadratic equation of the form ax² + bx + c = 0

Let us suppose the two consecutive numbers to be x and x + 1

According to the given condition x(x + 1) = 306

x² + x - 306 = 0, which is the required quadratic equation of the form ax² + bx + c = 0

Let age of Rohan = x

⇒ Rohan’s mother age = x + 26

After 3 years, Rohan’s age = x + 3

and mother’s age = (x + 29) years

According to the given condition

(x + 3) (x + 29) = 360

x² + 32x - 273 = 0, which is quadratic equation of the form ax² + bx + c = 0

x² + 39x - 7x -273 = 0

x(x + 39) -7 (x+ 39) = 0

(x-7)(x+39)= 0
x = 7 or x = -39

but age cannot be negative.

hence, x = 7

Let the speed of the train be x

Time taken to cover 480 km =

If the speed had been 8 km/h less than the train would take 3 hours to cover the same distance

Then speed will be x - 8 and time will be =

Speed =

⇒

⇒ 480x - 3x² - 3840 + 24x = 480x

⇒ - 3(x² - 8x + 1280) = 0

⇒ (x² - 8x + 1280) = 0

which is the required quadratic equation of the form ax² + bx + c = 0

x² - 3x - 10 = 0

⇒ x² - 5x + 2x - 10 = 0

⇒ x(x - 5) + 2(x - 5) = 0

⇒ (x + 2) (x - 5) = 0

The roots of the equation are those value for which (x + 2) (x - 5) = 0

⇒ x = - 2 or 5

- 2 and 5 are the roots of the given equation.

2x² + x - 6 = 0

⇒ 2x² + 4x - 3x - 6 = 0

⇒ 2x(x + 2) – 3(x + 2) = 0

⇒ (2x - 3)(x + 2) = 0

The roots of the equation are those value for which (x + 2) (2x - 3) = 0

⇒

⇒ = 0

⇒

⇒

The roots of the equation are those value for which

⇒

⇒

⇒

⇒

The roots of the equation are those value for which

⇒ x = 1/4

Hence 1/4 and 1/4 are the roots of the given equation.

100x² - 20x + 1 = 0

⇒ 100x² - 10x - 10x + 1 = 0

⇒ 10x(10x - 1) - 1( 10x - 1) = 0

⇒ (10x - 1) (10x - 1) = 0

The roots of the equation are those value for which (10x - 1) (10x - 1) = 0

⇒

.

x(x + 4) = 12

⇒ x² + 4x - 12 = 0

⇒ x² + 6x - 2x - 12 = 0

⇒ x( x + 6) - 2( x + 6) = 0

⇒ (x - 2) (x + 6) = 0

The roots of the equation are those value for which (x - 2) (x + 6) = 0

⇒ x = 2 or - 6

2 and - 6 are the roots of the given equation.

3x² - 5x + 2 = 0

⇒ 3x² – 2x - 3x + 2 = 0

⇒ x(3x - 2) - 1(3x - 2) = 0

⇒ (x - 1) (3x - 2) = 0

The roots of the equation are those value for which (x - 1) (3x - 2) = 0

⇒

Hence are the roots of the given equation.

⇒ x² - 2x - 3 = 0

⇒ x² – 3x + 1x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1) (x - 3) = 0

The roots of the equation are those value for which (x + 1) (x - 3) = 0

⇒ x = - 1 or 3

Hence - 1 and 3 are the roots of the given equation.

3(x - 4)² - 5(x - 4) = 12

⇒ 3( x² + 16 - 8x) - 5x + 20 - 12 = 0

⇒ 3x² + 48 – 24x - 5x + 20 - 12 = 0

⇒ 3x² – 29x + 56 = 0

⇒ 3x² – 21x - 8x + 56 = 0

⇒ 3x(x - 7) - 8( x - 7) = 0

⇒ (3x - 8) (x - 7) = 0

The roots of the equation are those value for which (3x - 8) (x - 7) = 0

⇒

are the roots of the given equation.

let the two numbers be x and y

According to the given condition

x + y = 27 ⇒ y = 27 - x

And x × y = 182

⇒ x (27 –x) = 182

⇒ 27x – x² - 182 = 0

⇒ x² – 27x + 182 = 0

⇒ x² – 13x – 14x + 182 = 0

⇒ x (x - 13) - 14 (x - 13) = 0

⇒ (x - 14) (x - 13) = 0

The roots of the equation are those value for which(x - 14) (x - 13) = 0

⇒ x = 14 or 13

Hence two numbers are 14 and 13

Let the two consecutive positive integers be x and x + 1

According to the given condition

x² + (x + 1)² = 613

⇒ x² + x² + 1 + 2x = 613

⇒ 2x² + 2x – 612 = 0

⇒ x² + x - 306 = 0

⇒ x² + 18x – 17x - 306 = 0

⇒ x(x + 18) - 17(x + 18) = 0

⇒ (x + 18)(x - 17) = 0

The roots of the equation are those value for which(x + 18)(x - 17) = 0

⇒ x = 17 or - 18

Hence the two positive integers are 17 and 18 or - 17 and - 18

Consider a right angle ΔABC

Let the base of the Δ be x

⇒ Altitude = x - 7

In the right-angled triangle

According to the Pythagoras theorem

(Hypotenuse)² = (base)² + (altitude)²

⇒ 13² = x² + (x - 7)²

⇒ 169 – x² – x² - 49 + 14x = 0

⇒ 2x² - 14x – 120 = 0

⇒ x² – 7x - 60 = 0

⇒ x² – 12x + 5x - 60 = 0

⇒ x( x - 12) + 5( x – 12) = 0

⇒ (x + 5) (x - 12) = 0

⇒ x = - 5 or 12

But base cannot be negative, so x = 12 cm

⇒ altitude = 12 - 7 = 5 cm

Hence 5 cm and 12 cm are the two sides of the given triangle.

Let x be the number of articles produced on that day

Thus cost of production of each article = 2x + 3

Total cost = Rs 90

According to the given condition

x(2x + 3) = 90

⇒ 2x² + 3x - 90 = 0

⇒ 2x² - 12x + 15x – 90 = 0

⇒ 2x(x - 6) + 15(x - 6) = 0

⇒ (2x + 15) (x - 6) = 0

⇒

Since the number of articles produced cannot be negative ∴ x = 6

And the cost of each article = 2 × 6 + 3 = Rs 15

Number of articles = 6

Cost of each article = 15

Let the length and the breath of the rectangle be l and b respectively.

According to the given condition

l × b = 40 m² ( area = length × breath)…………..1

and

Perimeter = 2(l + b)

2(l + b) = 28

⇒ l + b = 14

⇒ b = 14 –l

Putting value of b in 1

l( 14 - l) = 40

⇒ l² – 14 l + 40 = 0

⇒ l² – 10l – 4l + 40 = 0

⇒ l( l - 10) – 4(l - 10) = 0

⇒ (l - 4) ( l - 10) = 0

⇒ l = 4 or 10 m

Hence length is 4 m

And breath is 10 m.

let the base of the triangle be x cm and altitude be y cm

Given base = 4 + y

Area of the triangle = 1/2 × base × altitude

48 = 1/2 × (4 + y) × y

⇒ 96 = 4y + y²

⇒ y² + 4y – 96 = 0

⇒ y² + 12y - 8 y – 96 = 0

⇒ y( y + 12) - 8 ( y + 12) = 0

⇒ (y - 8) ( y + 12) = 0

⇒ y = 8 or - 12

Since altitude cannot be negative ∴ y = 8

⇒ base = 4 + 8 = 12 cm

Hence, Base = 12 cm; Altitude = 8 cm

Let x km /hour be the speed of the first train

Then the speed of the second train is (x - 5) km/hour

Given that they are 50 km apart after 2 hours

As per the given conditions

The first train travels from O to A and

The 2^nd train travels from O to B

Also given that AB = 50 Km

Since the OAB is a right angled triangle

∴ By Pythagoras theorem

OA² + OB² = AB²………….1

Now distance = speed × time

⇒ OA = x × 2

Also OB = 2(x - 5)

Putting value of OB and OA in 1

⇒ 8 x² – 40x + 100 = 2500

⇒ x² – 5x – 300 = 0

⇒ x(x - 20) + 15(x - 20) = 0

⇒ (x + 15) (x - 20) = 0

⇒ x = - 15 or 20

Since speed cannot be negative ∴ the speed of the 1^st train = 20 km/hour

And speed of 2^nd train is 15 km/hour.

Let the number of girls be x and number of boys be y

Given that the total number of students = 60

⇒ x + y = 60 ⇒ y = 60 –x

Total money collected = 1600

According to the given condition

1600 = xy + yx

⇒ 1600 = 2 xy

⇒ xy = 800

⇒ x( 60 –x) = 800

⇒ x² - 60x + 800 = 0

⇒ x² – 20x – 40x + 800 = 0

⇒ x ( x – 20) – 40 (x – 20) = 0

⇒ (x - 40) (x - 20) = 0

⇒ x = 40 or 20

Hence if the number of girls is 40, then the number of boys is 20

And if the number of girls is 20, then the number of boys is 40.

let the speed of the stream be x km/hour

∴ The speed of the boat upstream = (x - 3) km/hr and the speed of the boat downstream = (x + 3)km/hr

Similarly,

According to the given condition

⇒ 24(x + 3) + 24 (x - 3) = 6 (x + 3) (x - 3)

⇒ 48x = 6x² – 54

⇒ x² - 8x - 9 = 0

⇒ x² - 9x + x – 9 = 0

⇒ x( x - 9) + (x - 9) = 0

⇒ (x + 1) (x - 9) = 0

⇒ x = - 1 or 9

Since the speed of the stream cannot be negative so x = 9 km/hr

Hence the speed of the stream is 9 km/hr.

2x² + x - 4 = 0

The equation is of the form ax² + bx + c = 0

Since a≠0 ,so it exists

We use the quadratic formula to evaluate the roots

The two roots are

4x² + 4√3 + 3 = 0

The equation is of the form ax² + bx + c = 0

Since a≠0 ,so it exists

We use the quadratic formula to evaluate the roots

The two roots are

5x² - 7x - 6 = 0

The equation is of the form ax² + bx + c = 0

Since a≠0 ,so it exists

We use the method of factorization to solve it

⇒ 5x² - 10x + 3x - 6 = 0

⇒ 5x(x - 2) + 3(x - 2) = 0

⇒ (5x + 3)(x - 2) = 0

⇒ x

x² - 6x + 5 = 0

The equation is of the form ax² + bx + c = 0

Since a≠0 ,so it exists

We use the method of factorization to solve it

⇒ x² - 5x - x + 5 = 0

⇒ x(x - 5) - 1(x - 5) = 0

⇒ (x - 1)(x - 5) = 0

⇒ x = 1,5

We use the quadratic formula to evaluate the roots

The two roots are

We use the quadratic formula to evaluate the roots

The two roots are

We use the method of factorization to solve it

⇒ 5x² - 10x + 3x - 6 = 0

⇒ 5x(x - 2) + 3(x - 2) = 0

⇒ (5x + 3)(x - 2) = 0

⇒ x

We use the method of factorization to solve it

⇒ x² - 5x - x + 5 = 0

⇒ x(x - 5) - 1(x - 5) = 0

⇒ (x - 1)(x - 5) = 0

⇒ x = 1,5

⇒ x² - 3x - 1 = 0

We use the quadratic formula to evaluate the roots

The two roots are

⇒ 11(x² - 3x - 28) = - 330

⇒ x² - 3x - 28 = - 30

⇒ x² - 3x + 2 = 0

We use the method of factorization to solve

⇒ x² - 2x - x + 2 = 0

⇒ x(x - 2) - 1(x - 2) = 0

⇒ (x - 1)(x - 2) = 0

⇒ x = 1, 2

Let Present age of Rehman be x

Three Years ago his age was = x - 3

Five years from now his age will be = x + 5

According to the problem:

⇒ 6(x + 1) = (x - 3)(x + 5)

⇒ 6x + 6 = x² + 2x - 15

⇒ x² - 4x - 21 = 0

⇒ x² - 7x + 3x - 21 = 0

⇒ x(x - 7) + 3(x - 7) = 0

⇒ (x + 3)(x - 7) = 0

⇒ x = - 3 ,7

Since age cannot be negative so ,

Present age is 7 years

Maths = 12, English = 18 (or)

Maths = 13, English = 17

Let marks in Mathematics be x , marks in English be 30 - x

When marks in Mathematics is increased by 2 it becomes x + 2

When marks in English is decreased by 3 it becomes 27 - x

According to the problem the product of the two marks is 210

(x + 2)(27 - x) = 210

⇒ - x² + 25x + 54 = 210

⇒ x² - 25x + 156 = 0

Performing factorization we get:

⇒ x² - 13x - 12x + 156 = 0

⇒ x(x - 13) - 12(x - 13) = 0

⇒ (x - 12)(x - 13) = 0

⇒ x = 12,13

There will be two answers

Mathematics = 12 ,English = (30 - 12) = 18

Mathematics = 13, English = (30 - 13) = 17

120 m; 90 m

Let the shorter side be x

Longer side = x + 30

Diagonal = x + 60

Applying Pythagoras theorem

Diagonal² = Longer side² + shorter side²

⇒ (x + 60)² = (x + 30)² + x²

⇒ x² + 120x + 3600 = x² + 60x + 900 + x²

⇒ x² - 60x - 2700 = 0

Performing factorization we get:

⇒ x² - 90x + 30x - 2700 = 0

⇒ x(x - 90) + 30(x - 90) = 0

⇒ (x + 30)(x - 90) = 0

⇒ x = - 30, 90

Neglecting the negative term

Shorter Side = 90 m

Longer side = (90 + 30) = 120 m

18, 12; - 18, - 12

Let the larger number be x and smaller number be y

x² – y² = 180 …Equation (i)

Since the square of the smaller number is 8 times the larger number, so

⇒ x² – 8x - 180 = 0

Performing factorization we get:

⇒ x² - 18x + 10x - 180 = 0

⇒ x(x - 18) + 10(x - 18) = 0

⇒ (x + 10)(x - 18) = 0

x = - 10,18

Rearranging Equation (i)

y² = x² - 180

Putting x = - 10 we get

y² = - 80

Since the square is negative so the solution is neglected

Putting x = 18 we get

y² = 144

⇒ y = 12

Answer: 18,12 ;18, - 12

Let the speed of the train be x

Distance = 360 km

Time taken for a speed of x km/h

When speed is (x + 5) km/h time taken

According to the problem:

⇒ 360(x + 5) = 360x + x(x + 5)

⇒ x² + 5x - 1800 = 0

Performing factorization we get:

x² + 45x - 40x - 1800 = 0

⇒ x(x + 45) - 40(x + 45)

⇒ (x - 40)(x + 45) = 0

x = 40, - 45

Since Speed cannot be negative, so

Speed = 40 km/h

Let time taken by the tap of larger diameter be t and smaller diameter be t + 10

Part of tank filled in 1 hour by the tap of larger diameter

Part of tank filled in 1 hour by the tap of smaller diameter

Part of tank filled in 1 hour when both the taps were working together

So we can say:

⇒ 150t + 750 = 8(t² + 10t)

⇒ 8t² - 70t - 750 = 0

⇒ 4t² - 35t - 375 = 0

Performing factorization we get:

⇒ 4t² - 60t + 25t - 375 = 0

⇒ 4t(t - 15) + 25(t - 15) = 0

⇒ (4t + 25)(t - 15) = 0

t

Since time is always a positive quantity, so

Time taken by the tap of larger diameter = 15 h

Time taken by the tap of smaller diameter = (15 + 10) = 25 h

Let speed of passenger train be x and express train be x + 11

Distance travelled = 132 km

Time difference between the two trains = 1 hours

⇒ 132x + 1452 - 132x = x(x + 11)

⇒ x² + 11x - 1452 = 0

Performing factorization we get:

⇒ x² + 44x - 33x - 1452 = 0

⇒ x(x + 44) - 33(x + 44) = 0

⇒ (x - 33)(x + 44) = 0

⇒ x = 33, - 44

Since speed cannot be negative so

Speed of Passenger train = 33 kmph

Speed of Express Train = (33 + 11) = 44 kmph

Let the side of one square be x, another square be (x + 6)

Sum of Areas = 468 m²

x² + (x + 6)² = 468

2x² + 12x - 432 = 0

⇒ x² + 6x - 216 = 0

Performing factorization we get:

⇒ x² + 18x - 12x - 216 = 0

⇒ x(x + 18) - 12(x + 18) = 0

⇒ (x - 12)(x + 18) = 0

⇒ x = 12, - 18

Neglecting the negative value , we get

Side of squares = 12 m ,18 m

s = 96 + 80t - 4.9t²

When the ball reaches the ground, s = 0

4.9t² - 80t - 96 = 0

Since time cannot be negative so negative root is neglected

⇒ Time taken = 17.44 s

No. of diagonals =

n(n - 3) = 2 × 65

⇒ n² - 3n = 130

⇒ n² - 3n - 130 = 0

Performing factorization we get:

⇒ n² - 13n + 10n - 130 = 0

⇒ n(n - 13) + 10(n - 13) = 0

⇒ (n + 10)(n - 13) = 0

n = 13, - 10

Since no. of sides cannot be negative so

No. of Sides = 13

When No. of Diagonals is 50

n(n - 3) = 50 × 2

⇒ n² - 3n - 150 = 0

Discriminant = (9 - 4 × 1 × ( - 150)) = 609

Since 609 is not a perfect square so n can never be a whole number.

Hence 50 diagonals are not possible

2x² - 3x + 5 = 0

Discriminant = ( - 3)² - 4 × 2 × 5 = - 31

Since The Discriminant is negative so the roots are imaginary

3x² - 4√3 x + 4 = 0

Discriminant = ( - 4√3)² - 4 × 3 × 4 = 0

Since The Discriminant is zero so the roots are equal

The roots are

2x² - 6x + 3 = 0

Discriminant(D) = ( - 6)² - 4 × 2 × 3 = 12

Since The Discriminant is positive so the roots are real and distinct

The roots are evaluated by using the formula

The roots are

2x² + kx + 3 = 0

If the quadratic Equation has equal root then their Discriminant = 0

k² - 4 × 2 × 3 = 0

⇒ k² = 24

⇒ k = ±2√6

kx(x - 2) + 6 = 0

⇒ kx² - 2kx + 6 = 0

If the quadratic Equation has equal root then their Discriminant = 0

4k² - 4 × k × 6 = 0

⇒ k² - 6k = 0

⇒ k(k - 6) = 0

k = 0,6

If k = 0 then the equation is no longer Quadratic

Answer : k = 6

Let Breadth be x and length be 2x

Area = (Length × Breadth) = 800m²

2x² = 800

⇒ x² = 400

Discriminant = 4 × 1 × 400 = 1600

Hence it is possible

⇒ x = ±20

Neglecting the negative term we have,

Breadth = 20m

Length = 40m

Let age of one friend be x ,other 20 - x

Four years ago there age was (x - 4) ,(16 - x)

Product of age four years ago = 48

So we can say,

(x - 4)(16 - x) = 48

⇒ x² - 20x + 112 = 0

Discriminant = (- 20)² - 4 × 1 × 112 = - 48

Since the discriminant is negative so the roots of the equation are imaginary.

Answer: The above situation is not possible

Let Length of rectangular park be x and breadth be (40 - x)

Area = 400 m²

⇒ x(40 - x) = 400

⇒ x² - 40x + 400 = 0

Discriminant = ( - 40)² - 4 × 1 × 400

⇒ Discriminant = 1600 - 1600 = 0

Since Discriminant is 0 so it is possible

x² - 40x + 400 = 0

⇒ (x - 20)² = 0

⇒ x = 20 m

Answer: Length = 20m, Breadth = 20m