Quadratic Equations

x + = 2

⇒ = 2

⇒ x² + 1 = 2x

⇒ x² – 2x + 1 = 0

Comparing equation x² – 2x + 1 = 0 with ax² + bx + c = 0 we get

a = 1, b = – 2 and c = 1

Here a, b, c ∈ R and a ≠ 0

The degree of polynomial x² – 2x + 1 = 0 is 2

Hence x + = 2 is a quadratic equation

⇒ x² + 3x – 2x – 6 = 0

⇒ x² + x – 6 = 0

Comparing equation x² + x – 6 = 0 with ax² + bx + c = 0 we get

a = 1, b = 1 and c = – 6

Here a, b, c ∈ R and a ≠ 0

The degree of polynomial x² + x – 6 = 0 is 2

Hence (x — 2)(x + 3) = 0 is a quadratic equation

2x² – √5x + 2 = 0

Comparing equation 2x² – √5x + 2 = 0 with ax² + bx + c = 0 we get

a = 2, b = – √5 and c = 2

Here a, b, c ∈ R and a ≠ 0

The degree of polynomial 2x² – √5x + 2 = 0 is 2

Hence 2x² – √5x + 2 = 0 is a quadratic equation

For simplification of denominator use formula (a + b)(a – b) = a² – b²

⇒ – 2 = 3 × (x² – 1)

⇒ – 2 = 3x² – 3

⇒ 3x² – 3 + 2 = 0

⇒ 3x² + 0x – 1 = 0

Comparing equation 3x² + 0x – 1 = 0 with ax² + bx + c = 0 we get

a = 3, b = 0 and c = – 1

Here a, b, c ∈ R and a ≠ 0

The degree of polynomial 3x² + 0x – 1 = 0 is 2

Hence is a quadratic equation

For L.H.S use formula (a + b)(a – b) = a² – b²

⇒ 4x² – 1 = 4x² – 20x + 3x – 15

⇒– 1 = – 17x – 15

⇒ – 1 + 17x + 15 = 0

⇒ 17x + 14 = 0

⇒ 0x² + 17x + 14 = 0

Comparing equation 0x² + 17x + 14 = 0 with ax² + bx + c = 0 we get

a = 0, b = 17 and c = 14

Here a, b, c ∈ R but a = 0

Therefore, the degree of polynomial 0x² + 17x + 14 = 0 becomes 1

Hence (2x + 1)(2x — 1) = (4x + 3)(x — 5) is not a quadratic equation

Expand numerator using formula (a + b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b² and denominator using formula (a + b)(a – b) = a² – b²

⇒ – 4x = x² – 1

⇒ x² + 4x – 1 = 0

Comparing equation x² + 4x – 1 = 0 with ax² + bx + c = 0 we get

a = 1, b = 4 and c = – 1

Here a, b, c ∈ R and a ≠ 0

The degree of polynomial x² + 4x – 1 = 0 is 2

Hence – = is a quadratic equation

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

Where a = (2x + 3) and b = (3x + 2)

⇒ [(2x + 3) + (3x + 2)] × [(2x + 3) – (3x + 2)] = 13

⇒ (5x + 5) × [2x + 3 – 3x – 2] = 13

⇒ (5x + 5) × (1 – x) = 13

⇒ 5x – 5x² + 5 – 5x = 13

⇒ – 5x² + 5 – 13 = 0

⇒ – 5x² + 0x – 8 = 0

Comparing equation – 5x² + 0x – 8 = 0 with ax² + bx + c = 0 we get

a = – 5, b = 0 and c = – 8

Here a, b, c ∈ R and a ≠ 0

The degree of polynomial – 5x² + 0x – 8 = 0 is 2

Hence (2x + 3)² — (3x + 2)² = 13 is a quadratic equation

x = 2 will be solution of quadratic equation x² — 3x + 2 = 0 if x = 2 satisfies the given equation

put x = 2 in L.H.S of x² — 3x + 2 = 0

⇒ L.H.S = 2² – 3 × 2 + 2

⇒ L.H.S = 4 – 6 + 2

⇒ L.H.S = – 2 + 2

⇒ L.H.S = 0

⇒ L.H.S = R.H.S

Thus x = 2 satisfies the equation x² — 3x + 2 = 0 hence x = 2 is a solution

x = 2 will be solution of quadratic equation x² + x — 2 = 0 if x = 2 satisfies the given equation

put x = 2 in L.H.S of x² + x — 2 = 0

⇒ L.H.S = 2² – 2 + 2

⇒ L.H.S = 4

⇒ L.H.S ≠ R.H.S

Thus x = 2 does not satisfies the equation x² + x — 2 = 0 hence x = 2 is not a solution

x = 1 will be solution of quadratic equation

if x = 1 satisfies the given equation

put x = 1 in L.H.S of 1/(3x + 1) – 1/(2x – 1) + 3/4 = 0

⇒ L.H.S

⇒ L.H.S

⇒ L.H.S

⇒ L.H.S

⇒ L.H.S = 0

⇒ L.H.S = R.H.S

Thus x = 1 satisfies the equation hence x = 1 is a solution

will be solution of quadratic equation (3x — 8)(2x + 5) = 0 if satisfies the given equation

put in L.H.S of (3x — 8)(2x + 5) = 0

⇒ L.H.S

⇒ L.H.S ]

⇒ L.H.S

⇒ L.H.S = 0

⇒ L.H.S = R.H.S

Thus satisfies the equation (3x — 8)(2x + 5) = 0 hence is a solution

As x = 1 is a root of ax² + bx + c = 0 hence x = 1 will satisfy the equation ax² + bx + c = 0

Put x = 1 in ax² + bx + c = 0

⇒ a × (1)² + b × (1) + c = 0

⇒ a + b + c = 0

Hence proved

As x = – 1 is a root of x² — px + q = 0 hence x = – 1 will satisfy the equation x² — px + q = 0

Put x = – 1 in x² — px + q = 0

⇒ (– 1)² – p × (– 1) + q = 0

⇒ 1 + p + q = 0

⇒ p + q + 1 = 0

Hence proved

One of the root of equation x² — kx + 6 = 0 is 3 which means x = 3 is a root

As x = 3 is a root of x² — kx + 6 = 0 hence x = 3 will satisfy the equation x² — kx + 6 = 0

Put x = 3 in equation x² — kx + 6 = 0

⇒ 3² – k × (3) + 6 = 0

⇒ 9 – 3k + 6 = 0

⇒ 15 – 3k = 0

⇒ 15 = 3k

⇒ k =

⇒ k = 3

One of the root of equation x² + 3(k + 2)x — 9 = 0 is – 3 which means x = – 3 is a root

As x = – 3 is a root of x² + 3(k + 2)x — 9 = 0 hence x = – 3 will satisfy the equation x² + 3(k + 2)x — 9 = 0

Put x = – 3 in equation x² + 3(k + 2)x — 9 = 0

⇒ (– 3)² + [3 × (k + 2) × (– 3)] – 9 = 0

⇒ 9 + (– 9) × (k + 2) – 9 = 0

⇒ (– 9) × (k + 2) = 0

⇒ k + 2 = 0

⇒ k = – 2

27 and 48 are divisible by 3 take factor 3 common

⇒ 3 × (9x² – 16) = 0

⇒ 9x² – 16 = 0

⇒ (3x)² – 4² = 0

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

Where a = 3x and b = 4

⇒ (3x + 4)(3x – 4) = 0

⇒ 3x + 4 = 0 or 3x – 4 = 0

⇒ 3x = – 4 or 3x = 4

Therefore and

(x — 7)² — 4² = 0

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

Where a = (x – 7) and b = 4

⇒ (x – 7 + 4)(x – 7 – 4) = 0

⇒ (x – 3)(x – 11) = 0

⇒ x – 3 = 0 or x – 11 = 0

Therefore x = 3 and x = 11

6x² + 13x + 6 = 0

⇒ 6x² + 9x + 4x + 6 = 0

taking 3x common from first two terms and 2 common from next two

⇒ 3x(2x + 3) + 2(2x + 3) = 0

⇒ (3x + 2)(2x + 3) = 0

⇒ 3x + 2 = 0 or 2x + 3 = 0

⇒ 3x = – 2 or 2x = – 3

Therefore and

15x² — 16x + 1 = 0

⇒ 15x² – 15x– x + 1 = 0

taking 15x common from first two terms and – 1 common from next two

⇒ 15x(x – 1) – 1(x – 1) = 0

⇒ (15x – 1)(x – 1) = 0

⇒ 15x – 1 = 0 or x – 1 = 0

⇒ 15x = 1 or x = 1

Therefore and x = 1

√5x² – 4x – √5 = 0

⇒ √5x² – 5x + x – √5 = 0

taking √5x common from first two terms and 1 common from next two

⇒ √5x(x – √5) + 1(x – √5) = 0

⇒ (√5x + 1)(x – √5) = 0

⇒ √5x + 1 = 0 or x – √5 = 0

⇒ √5x = – 1 or x = √5

Therefore and x = √5

⇒ 6(x² + 1) = 13x

⇒ 6x² + 6 = 13x

⇒ 6x² – 13x + 6 = 0

⇒ 6x² – 9x– 4x + 6 = 0

taking 3x common from first two terms and – 2 common from next two

⇒ 3x(2x – 3) – 2(2x – 3) = 0

⇒ (3x – 2)(2x – 3) = 0

⇒ 3x – 2 = 0 or 2x – 3 = 0

⇒ 3x = 2 or 2x = 3

Therefore and

Comparing equation 6x² — 13x + 6 = 0 with ax² + bx + c = 0 we get

a = 6, b = – 13 and c = 6

Discriminant (D) = b² – 4ac

⇒ D = (– 13)² – 4(6)(6)

⇒ D = 169 – 4 × 36

⇒ D = 169 – 144

⇒ D = 25

As D > 0 roots of equation 6x² — 13x + 6 = 0 are real and distinct

√6x² – 5x + √6 = 0

Comparing equation √6x² – 5x + √6 = 0 with ax² + bx + c = 0 we get

a = √6, b = – 5 and c = √6

Discriminant (D) = b² – 4ac

⇒ D = (– 5)² – 4(√6)(√6)

⇒ D = 25 – 4 × 6

⇒ D = 25 – 24

⇒ D = 1

As D > 0 roots of equation √6x² – 5x + √6 = 0 are real and distinct

Comparing equation 24x² — 17x + 3 = 0 with ax² + bx + c = 0 we get

a = 24, b = – 17 and c = 3

Discriminant (D) = b² – 4ac

⇒ D = (– 17)² – 4(24)(3)

⇒ D = 289 – 4 × 72

⇒ D = 289 – 288

⇒ D = 1

As D > 0 roots of equation 24x² — 17x + 3 = 0 are real and distinct

Comparing equation x² + 2x + 4 = 0 with ax² + bx + c = 0 we get

a = 1, b = 2 and c = 4

Discriminant (D) = b² – 4ac

⇒ D = 2² – 4(1)(4)

⇒ D = 4 – 16

⇒ D = – 12

As D < 0 equation x² + 2x + 4 = 0 has no real solution

Comparing equation x² + x + 1 = 0 with ax² + bx + c = 0 we get

a = 1, b = 1 and c = 1

Discriminant (D) = b² – 4ac

⇒ D = 1² – 4(1)(1)

⇒ D = 1 – 4

⇒ D = – 3

As D < 0 equation x² + x + 1 = 0 has no real solution

Comparing equation x² — 3√3x — 30 = 0 with ax² + bx + c = 0 we get

a = 1, b = – 3√3 and c = – 30

Discriminant (D) = b² – 4ac

⇒ D = (– 3√3)² – 4(1)(– 30)

⇒ D = (3²)(√3)² + 120

⇒ D = 9 × 3 + 120

⇒ D = 27 + 120

⇒ D = 147

As D > 0 roots of equation x² — 3√3x — 30 = 0 are real and distinct

For roots to be real and distinct D should be greater than 0

Discriminant (D) = b² – 4ac

In the discriminant b² is a positive number since square cannot be negative given a > 0 which means a is also positive

Now consider the product – 4 × c

As c < 0 c is negative

We are multiplying two negative numbers which would result in a positive number which means – 4 × c is also positive

So, we can conclude that b² – 4ac is positive

Hence D > 0

Hence roots are real and distinct

Comparing equation x² — (3k — 2)x + 2k = 0 with ax² + bx + c = 0 we get

a = 1, b = – (3k – 2) and c = 2k

Discriminant (D) = b² – 4ac

As roots are real and equal D = 0

⇒ b² – 4ac = 0

⇒ [ – (3k – 2)]² – 4(1)(2k) = 0

⇒ (3k – 2)² – 8k = 0

Expand using (a – b)² = a² – 2ab + b²

⇒ 9k² – 12k + 4 – 8k = 0

⇒ 9k² – 18k– 2k + 4 = 0

taking 9k common from first two terms and – 2 common from next two

⇒ 9k(k – 2) – 2(k – 2) = 0

⇒ (9k – 2)(k – 2) = 0

⇒ 9k – 2 = 0 or k – 2 = 0

⇒ 9k = 2 or k = 2

Therefore k = and k = 2

Comparing equation (k + 1)x² — 2(k—1)x + 1 = 0 with ax² + bx + c = 0 we get

a = (k + 1), b = – 2(k – 1) and c = 1

Discriminant (D) = b² – 4ac

As roots are real and equal D = 0

⇒ b² – 4ac = 0

⇒ [ – 2(k – 1)]² – 4(k + 1)(1) = 0

⇒ (4)(k – 1)² – 4k – 4 = 0

Expand (k – 1)² using (a – b)² = a² – 2ab + b²

⇒ 4(k² – 2k + 1) – 4k – 4 = 0

⇒ 4k² – 8k – 4k = 0

⇒ 4k² – 12k = 0

Take 4k factor common

⇒ 4k(k – 3) = 0

⇒ 4k = 0 or k – 3 = 0

⇒ k = 0 or k = 3

Therefore k = 0 and k = 3

(To avoid confusion between variables here I have taken standard from as px² + qx + r = 0 and accordingly the discriminant)

Comparing equation ax² + 2bx + c = 0 with standard form px² + qx + r = 0 we get

p = a, q = 2b and r = c

Discriminant (D) = q² – 4pq

As roots are real and equal D = 0

⇒ q² – 4pq = 0

⇒ (2b)² – 4(a)(c) = 0

⇒ 4b² – 4ac = 0

⇒ 4b² = 4ac

⇒ b² = ac

⇒ b × b = a × c

⇒ =

⇒ b:c = a:b

⇒ a:b = b:c

Hence proved

Comparing equation x² + 10x + 6 = 0 with ax² + bx + c = 0 we get

a = 1, b = 10 and c = 6

Discriminant (D) = b² – 4ac

⇒ D = 10² – 4(1)(6)

⇒ D = 100 – 24

⇒ D = 76

D > 0 implies roots are real and distinct and given by

Therefore x = – 5 + √19 and x = – 5 – √19

Comparing equation x² + 5x — 1 = 0 with ax² + bx + c = 0 we get

a = 1, b = 5 and c = – 1

Discriminant (D) = b² – 4ac

⇒ D = 5² – 4(1)(– 1)

⇒ D = 25 + 4

⇒ D = 29

D > 0 implies roots are real and distinct and given by

Therefore and

Comparing equation x² — 3x — 2 = 0 with ax² + bx + c = 0 we get

a = 1, b = – 3 and c = – 2

Discriminant (D) = b² – 4ac

⇒ D = (– 3)² – 4(1)(– 2)

⇒ D = 9 + 8

⇒ D = 17

D > 0 implies roots are real and distinct and given by

Therefore and

Comparing equation x² — 3√6x + 12 = 0 with ax² + bx + c = 0 we get

a = 1, b = – 3√6 and c = 12

Discriminant (D) = b² – 4ac

⇒ D = (– 3√6)² – 4(1)(12)

⇒ D = (3²)(√6)² – 48

⇒ D = 9 × 6 – 48

⇒ D = 54 – 48

⇒ D = 6

D > 0 implies roots are real and distinct and given by

and

and

Therefore x = 2√6 and x = √6

Comparing equation 3x² + 5√2x + 2 = 0 with ax² + bx + c = 0 we get

a = 3, b = 5√2 and c = 2

Discriminant (D) = b² – 4ac

⇒ D = (5√2)² – 4(3)(2)

⇒ D = (5²)(√2)² – 4 × 6

⇒ D = 25 × 2 – 24

⇒ D = 50 – 24

⇒ D = 26

D > 0 implies roots are real and distinct and given by

Therefore and

Cross multiply

⇒ 5 × (x² – 1) = 4 × (x² + 1)

⇒ 5x² – 5 = 4x² + 4

⇒ 5x² – 5 – 4x² – 4 = 0

⇒ x² – 9 = 0

⇒ x² + 0x – 9 = 0

Comparing equation x² + 0x – 9 = 0 with ax² + bx + c = 0 we get

a = 1, b = 0 and c = – 9

Discriminant (D) = b² – 4ac

⇒ D = 0² – 4(1)(– 9)

⇒ D = 36

D > 0 implies roots are real and distinct and given by

and

Therefore x = 3 and x = – 3

Let the two numbers be x and 27 – x so that their sum is 27

Product = 182

⇒ x(27 – x) = 182

⇒ 27x – x² = 182

⇒ x² – 27x + 182 = 0

⇒ x² – 14x– 13x + 182 = 0

taking x common from first two terms and – 13 common from next two

⇒ x(x – 14) – 13(x – 14) = 0

⇒ (x – 13)(x – 14) = 0

⇒ (x – 13) = 0 or (x – 14) = 0

Therefore x = 13 and x = 14

Take any value of x 13 or 14 if we subtract it from 27 we get the other value 13/14

Hence the numbers are 13 and 14 whose sum is 27 and product is 182

Let the two consecutive numbers be x and (x + 1)

Sum of squares of these numbers is 365

⇒ x² + (x + 1)² = 365

⇒ x² + x² + 2x + 1 = 365

⇒ 2x² + 2x + 1 – 365 = 0

⇒ 2x² + 2x – 364 = 0

Take 2 as common factor

⇒ 2 × (x² + x – 182) = 0

⇒ x² + x – 182 = 0

⇒ x² + 14x – 13x – 182 = 0

taking x common from first two terms and – 13 common from next two

⇒ x(x + 14) – 13(x + 14) = 0

⇒ (x – 13)(x + 14) = 0

⇒ (x – 13) = 0 or (x + 14) = 0

Therefore x = 13 and x = – 14

if we take x = 13 then the two consecutive numbers whose sum of squares is 365 will be 13 and 14

if we take x = – 14 then the two consecutive numbers whose sum of squares is 365 will be – 13 and – 14

Let age of one friend be x and another friend be (20 – x) so that their sum is 20

Age four years ago

Age of one friend would have been (x – 4) and another friend would have been (20 – x – 4) which is (16 – x)

Given that four years ago the product of their ages was 48

⇒ (x – 4)(16 – x) = 48

⇒ 16x – x² – 64 + 4x = 48

⇒ – x² – 64 + 20x = 48

⇒ x² + 64 – 20x + 48 = 0

⇒ x² – 20x + 112 = 0

Comparing equation x² – 20x + 112 = 0 with ax² + bx + c = 0 we get

a = 1, b = – 20 and c = 112

Discriminant (D) = b² – 4ac

⇒ D = (– 20)² – 4(1)(112)

⇒ D = 400 – 448

⇒ D = –48

D<0 which means no real values of x

As no real values of x the equation which we formed using given statements cannot be true

Hence the statements cannot be true

Let the breadth of garden be b

Thus length = 2b

Area of garden = 800 m²

Area = length × breadth

⇒ 800 = 2b × b

⇒ 800 = 2b²

⇒ b² = 400

⇒ b² = 20²

⇒ b² – 20² = 0

Using identity (a + b)(a – b) = a² – b²

⇒ (b + 20)(b – 20) = 0

⇒ (b + 20) = 0 or (b – 20) = 0

Therefore b = 20 and not – 20 because b cannot be negative as b is breadth of rectangle and length or breadth cannot be negative

Breadth = 20

Therefore length = 2 × 20 = 40

Therefore, length of the garden is 40 m

Let l be length of rectangle and b be breadth of rectangular garden

Perimeter = 360 m

Perimeter = 2 × (length + breadth) = 2(l + b)

⇒ 2(l + b) = 360

⇒ l + b = 180

⇒ l = 180 – b …(i)

Area = 8000

Area = length × breadth

⇒ 8000 = lb

Using (i)

⇒ 8000 = (180 – b) × b

⇒ 8000 = 180b – b²

⇒ b² – 180b + 8000 = 0

⇒ b² – 100b– 80b + 8000 = 0

taking b common from first two terms and – 80 common from next two

⇒ b² – 100b – 80b + 8000 = 0

⇒ b(b – 100) – 80(b – 100) = 0

⇒ (b – 80)(b – 100) = 0

⇒ (b – 80) = 0 or (b – 100) = 0

b = 80 m because it is given that breadth is smaller than length

using (i) l = 180 – 80 = 100 m

Therefore, length of garden is 100 m and breadth is 80 m

Distance = 35 km

Let the usual speed be x

And t be the time taken to reach destination when speed is x

…(i)

If he travels at a speed (x + 2) km/hour time taken is (t – 2) hours

Distance is same 35 km

Speed =

Using (i)

⇒ (x + 2)(35 – 2x) = 35x

⇒– 2x² + 70 – 4x = 0

divide by – 2

⇒ x² + 2x – 35 = 0

⇒ x² + 7x– 5x – 35 = 0

taking x common from first two terms and – 5 common from next two

⇒ x(x + 7) – 5(x + 7) = 0

⇒ (x – 5)(x + 7) = 0

⇒ (x – 5) = 0 or (x + 7) = 0

Thus x = 5 and not 7 because x represent speed and speed cannot be negative

The usual speed of cyclist = 5 km/hour

Let the breadth of the ground be b

Diagonal = b + 60

Length = b + 30

Using Pythagoras theorem

(length)² + (breadth)² = (diagonal)²

⇒ (b + 30)² + b² = (b + 60)²

⇒ (b + 30)² – (b + 60)² = – b²

Using identity (a + b)(a – b) = a² – b²

⇒ (b + 30 + b + 60)(b + 30 – b – 60) = – b²

⇒ (2b + 90)(– 30) = – b²

⇒ b² – 60b – 2700 = 0

⇒ b² – 90b + 30b – 2700 = 0

taking b common from first two terms and 30 common from next two

⇒ b(b – 90) + 30(b – 90) = 0

⇒ (b + 30)(b – 90) = 0

⇒ (b + 30) = 0 or (b – 90) = 0

Thus b = 90 m as b cannot be negative because b represents breadth of rectangle

Length = b + 30 = 90 + 30 = 120 m

Area = length × breadth = 120 × 90 = 10800 m²

Therefore, area of ground is 10800 m²

Let the positive integers x, (x + 1) and (x + 2) be sides of right angled triangle

As (x + 2) will be the greatest number so (x + 2) is the hypotenuse

Using Pythagoras theorem

⇒ x² + (x + 1)² = (x + 2)²

⇒ x² = (x + 2)² – (x + 1)²

Using identity (a + b)(a – b) = a² – b²

⇒ x² = (x + 2 + x + 1)(x + 2 – x – 1)

⇒ x² = (2x + 3)(1)

⇒ x² = 2x + 3

⇒ x² – 2x – 3 = 0

⇒ x² – 3x + x – 3 = 0

taking x common from first two terms and 1 common from next two

⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x + 1)(x – 3) = 0

⇒ (x + 1) = 0 or (x – 3) = 0

Thus x = 3 because x cannot be negative since x represent he side of a triangle and side cannot be a negative quantity

x + 1 = 3 + 1 = 4

x + 2 = 3 + 2 = 5

Thus, the three sides are 3, 4 and 5

As it is a right angled triangle one side would be base and the other height

Base = 3 and height = 4

Area of triangle = × base × height

Area of triangle = × 3 × 4 = × 12

Therefore, area of triangle is 6 unit²

(x – 2√ 3)(x + 2√ 3) = 0

(∵ a² – b² = (a + b)(a – b), and 12 = (2√ 3)²)

⇒ x – 2√ 3 = 0 or x + 2√ 3 = 0

⇒ x = 2√ 3 or x = – 2√ 3

x² – 7x – 60 = 0

⇒ x² – (12 – 5)x – 60 = 0

⇒ x² – 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

⇒ x – 12 = 0 or x + 5 = 0

⇒ x = 12 or x = – 5

x²— (8 + 7)x + 56 = 0

⇒ x² – 8x – 7x + 56 = 0

⇒ x(x – 8) – 7(x – 8) = 0

⇒ (x – 7)(x – 8) = 0

⇒ x – 7 = 0 or x – 8 = 0

⇒ x = 7 or x = 8

(∵ (a² + b²) + (a² – b²) = 2(a² + b²)

⇒ 4(8x² + 18) = 17(4x² – 9)

⇒ 32x² + 72 = 68x² – 153

⇒ 68x² – 32x² = 153 – 72

⇒ 36x² = 81

⇒ (15x + 19)(x + 2) = (12x² + 64x + 20)

⇒ 15x² + 19x + 30x + 38 = 12x² + 64x + 20

⇒ 15x² + 49x + 38 = 12x² + 64x + 20

⇒ 15x² – 12x² + 49x – 64x + 38 – 20 = 0

⇒ 3x² – 15x + 18 = 0

⇒ 3x² – (18 – 3)x + 18 = 0

⇒ 3x² – 18x – 3x + 18 = 0

⇒ 3x(x – 6) – 3(x – 6) = 0

⇒ (3x – 3)(x – 6) = 0

⇒ x = 1 or x = 6.

x² — 24x + 144 – 160 = 0

⇒ (x – 12)² – 160 = 0

⇒ (x – 12)² – (4√ 10)² = 0

⇒ (x – 12 – 4√ 10)(x – 12 + 4√ 10) = 0

⇒ x – 12 – 4√ 10 = 0 or x – 12 + 4√ 10 = 0

⇒ x = 12 + 4√ 10 or x = 12 – 4√ 10

Multiplying the whole equation by 12, we get –

⇒ 36x² + 84x — 240 = 0

⇒ 36x² + 84x + 49 – 289 = 0

⇒ (6x + 7)² – 17² = 0

⇒ (6x + 7 – 17)(6x + 7 + 17) = 0

⇒ (6x – 10)(6x + 24) = 0

⇒6x – 10 = 0 or 6x + 24 = 0

x²— 2(1)(5)x + 5² = 0

⇒ (x – 5)² = 0

⇒ (x – 5)(x – 5) = 0

⇒ x – 5 = 0 or x – 5 = 0

⇒ x = 5

x² + x² + 10x + 25 – 625 = 0

⇒ 2x² + 10x – 600 = 0

Dividing the whole equation by 2, we get,

⇒ x² + 5x – 300 = 0

⇒(x + 20) = 0 or (x – 15) = 0

⇒ x = – 20 or x = 15

(x + 2)(x + 3) = 240

⇒ x(x + 3) + 2(x + 3) = 240

⇒ x² + 2x + 3x + 6 = 240

⇒ x² + 5x – 234 = 0

(ii) Take both the squares so formed and apply the formula a² - b² = (a + b) (a - b)

⇒(x – 18) = 0 or (x + 13) = 0

⇒ x = 18 or x = – 13 Answer..

Let the divided two parts be x and 20 – x.

⇒ x² + (20 – x)² = 218

⇒ x² + 400 – 40x + x² = 218

⇒ 2x² – 40x + 182 = 0

Dividing the whole equation by 2, we get –

⇒ x² – 20x + 91 = 0

⇒ x² – (7 + 13)x + 91 = 0

⇒ x² – 7x – 13x + 91 = 0

⇒ x(x – 7) – 13(x – 7) = 0

⇒ (x – 7)(x – 13) = 0

⇒ x – 7 = 0 or x – 13 = 0

⇒ x = 7 or x = 13

∴ the two parts are 7 and 13.

Let the usual speed be – x km/hr and time taken to cover 200 km be – t.

⇒

Now, when the car speed will be increased by the 10 km/hr, then the speed will become – (x + 10) km/hr.

Now, let the time taken now be T

But given that car took 1 hr. less to cover 200 km by new speed.

⇒ t – T = 1

⇒ 200x + 2000 – 200x = x² + 10x

⇒ x² + 10x – 2000 = 0

⇒ x² + (50 – 40)x – 2000 = 0

⇒ x² + 50x – 40x – 2000 = 0

⇒ x(x + 50) – 40(x + 50) = 0

⇒ (x + 50)(x – 40) = 0

⇒ x = – 50 or x = 40

And speed cannot be negative.

⇒ x = 40 km/hr.

Let the usual speed be – x km/hr and time taken to cover 400 km be – t.

Now, when the car speed will be decreased by the 5 km/hr, then the speed will become – (x – 5) km/hr.

Now, let the time taken now be T

But given that car took 4 hr. more to cover 200 km by new speed.

⇒ T = t + 4

⇒ 400x + 2000 – 400x = 4x² – 20x

⇒ 4x² – 20x – 2000 = 0

Dividing whole equation by 4, we get –

⇒ x² – 5x – 500 = 0

⇒ x² – (25 – 20)x – 2000 = 0

⇒ x² – 25x + 20x – 2000 = 0

⇒ x(x – 25) + 20(x – 25) = 0

⇒ (x – 25)(x + 20) = 0

⇒ x = 25 or x = – 20

And speed cannot be negative.

⇒ x = 25 km/hr.

Let speed of boat in still water be x km/hr.

∵ speed of river is given as 1 km/hr

⇒ speed of boat downstream = (x + 1) km/hr.

∴ let time taken to go 112 km downstream = t

Now, speed of boat upstream will be (x – 1) km/hr.

∴ let time taken to go 112 km upstream = T

Given that total journey time was 15 hr

⇒ t + T = 15

⇒ 112x – 112 + 112x + 112 = 15x² – 15

⇒ 15x² – 224x – 15 = 0

⇒ 15x² – 224x – 15 = 0

⇒ 15x² – (225 – 1)x – 15 = 0

⇒ 15x² – 225x + x – 15 = 0

⇒ 15x(x – 15) + 1(x – 15) = 0

o r

And speed cannot be negative.

⇒ x = 15 km/hr.

Let the no.be x.

⇒ 15x² + 15 = 34x

⇒ 15x² – 34x + 15 = 0

⇒ 15x² – (9 + 25)x + 15 = 0

⇒ 15x² – 9x – 25x + 15 = 0

⇒ 3x(5x – 3) – 5(5x – 3) = 0

⇒ (3x – 5) = 0 or (5x – 3 = 0)]]

or

and

∴ The required no is

Let the usual speed be – x km/hr and time taken to cover 400 km be – t.

⇒

Now, when the car speed will be increased by the 20km/hr, then the speed will become – (x + 20) km/hr.

Now, let the time taken now be T

But given that car took 1 hr. less to cover 200 km by new speed.

⇒ t – T = 1

⇒ 400x + 8000 – 400x = x² + 20x

⇒ x² + 20x – 8000 = 0

⇒ x² + (100 – 80)x – 2000 = 0

⇒ x² + 100x – 80x – 8000 = 0

⇒ x(x + 100) – 80(x + 100) = 0

⇒ (x + 100)(x – 80) = 0

⇒ x = – 1000 or x = 80

And speed cannot be negative.

⇒ x = 80 km/hr.

And speed of faster car = 100 km/hr.

Let the present age of Virat be x yrs.

⇒ His age 7 yrs ago was = (x – 7) yrs. and 7 years later will be = (x + 7) yrs.

⇒ (x – 7)(x + 7) = 480

⇒ x² – 49 = 480

⇒ x² – 49 – 480 = 0

⇒ x² – 529 = 0

⇒ x² – 23² = 0

⇒ (x + 23)(x – 23) = 0

⇒ x + 23 = 0 or x – 23 = 0

⇒ x = 23 or x = – 23

But age cannot be negative

⇒ x = 23 years.

Let present age of Sachin be x yrs.

∴ 8 yrs ago, his age was = (x – 8) yrs.

And, 2 yrs later his age will be = (x + 2) yrs.

⇒ (x – 8)(x + 2) = 1200

⇒ x² – 6x – 16 = 1200

⇒ x² – 6x – 1216 = 0

⇒ x² – (38 – 32)x – 1216 = 0

⇒ x² – 38x + 32x – 1216 = 0

⇒ x(x – 38) + 32(x – 38) = 0

⇒ (x – 38) + (x + 32) = 0

⇒ x – 38 = 0 or x + 32 = 0

⇒ x = 38 or x = – 32

But age cannot be negative.

⇒ x = 38 years.

Let present age of Anita be x yrs.

∵ sunita’s age is 2 yrs less than 6 times age of Anita.

⇒ Present age of Anita = (6x – 2) yrs.

∴ five yrs later age of Anita will be – (x + 5)yrs.

And age of Sunita = (6x – 2) + 5 yrs

= 6x + 3 years.

⇒ (x + 5)(6x + 3) = 330

⇒ 6x² + 33x + 15 = 330

⇒6x² + 33x – 315 = 0

Dividing the whole equation by 3, we get –

⇒2x² + 11x – 105 = 0

⇒2 + (21 – 10)x – 105 = 0

⇒2x² + 21x – 10x – 105 = 0

⇒x(2x + 21) – 5(2x + 21) = 0

⇒ (2x + 21)(x – 5) = 0

⇒ 2x + 21 = 0 or x – 5 = 0

But age cannot be negative

⇒ x = 5 years.

∴ present age of Anita = 5 yrs.

And present age of Sunita = 6x – 2

= 6(5) – 2

= 28 years.

And, ∵ Anita was born five years ago,

⇒ Age of Sunita at that time = 28 – 5 yrs.

= 23 yrs.

And, S = 325

⇒ n² + n = 325(2)

⇒ n² + n – 650 = 0

⇒ n² + (26 – 25)n – 650 = 0

⇒ n² + 26n – 25n – 650 = 0

⇒ n(n + 26) – 25(n + 26) = 0

⇒ (n – 25)(n + 26) = 0

⇒ n – 25 = 0 or n + 26 = 0

⇒ n = 25 or n = – 26

Given that n is a natural no.

⇒ n cannot be negative.

⇒ n = 25

Let length of shortest side(base) be x.

⇒ hypotenuse = 3x – 2.

⇒ remaining side(perpendicular) = 2x + 2

Now, According to Pythagoras theorem,

Hypotenuse² = Base² + Perpendicular²

⇒(3x – 2)² = x² + (2x + 2)²

⇒ 9x² + 4 – 12x = x² + 4x² + 4 + 8x

⇒ 4x² – 20x = 0

Dividing the whole equation by 4, we get –

⇒ x² – 5x = 0

⇒ x(x – 5) = 0

⇒ x = 0 or x – 5 = 0

But side cannot be zero

⇒ x(Base) = 5

⇒Perpendicular = 2x + 2

= 2(5) + 2

= 12

⇒ hypotenuse = 3x – 2

= 3(5) – 2

= 13

Now, area of right triangle –

= 30 sq. unit.

Let first odd positive integer be x.

⇒ second odd positive integer = x + 2

⇒ x² + (x + 2)² = 290

⇒ x² + x² + 4 + 4x = 290

⇒ 2x² + 4x – 286 = 0

Dividing the whole equation by 2, we get –

⇒ x² + 2x – 143 = 0

⇒ x² + (13 – 11)x – 143 = 0

⇒ x² + 13x – 11x – 143 = 0

⇒ x(x + 13) – 11(x + 13) = 0

⇒ (x – 11)(x + 13) = 0

⇒ x = 11 or x = – 13

But x is a positive integer

⇒ x = 11 is the first odd positive integer.

And, second consecutive odd positive integer = x + 2

= 13

Let two consecutive even natural numbers be x and x + 2.

⇒ x(x + 2) = 224

⇒ x² + 2x – 224 = 0

⇒ x² + (16 – 14)x – 224 = 0

⇒ x² + 16x – 14x – 224 = 0

⇒ x(x + 16) – 14(x + 16) = 0

⇒ (x – 14)(x + 16) = 0

⇒ x = 14 or x = – 16

But x is a natural no.

⇒ x = 14 is the first natural no.

And, second no. = x + 2

= 16

The product of digits of a two – digit number is 8

Let digit at tens place be x

Now,given –

⇒ x⁴ + 64 = 20x²

⇒ x⁴ – 20x² + 64 = 0

⇒ x⁴ – (16 + 4)x² + 64 = 0

⇒ x²(x² – 4) – 16(x² – 4) = 0

⇒ (x² – 16) (x² – 4) = 0

⇒ x = 4 or x = 2

For x = 4,

Original no.

= 42

For x = 2,

Original no.

= 24

Given no. is less than 25 –

⇒ two – digit no. = 24

Let price of 1 kg sugar be x.

Quantity of sugar

If price decreased by Rs. 5,then new price = x – 5

∴ quantity can be

Given, one can buy 1 kg more sugar –

⇒ 150x – 150x + 750 = x² – 5x

⇒ x² – 5x – 750 = 0

⇒ x² – (30 – 25)x – 750 = 0

⇒ x² – 30x + 25x – 750 = 0

⇒ x(x – 30) + 25(x – 30) = 0

⇒ (x – 30)(x + 25) = 0

⇒ x = 30 or x = – 25

But price cannot be negative.

⇒ x = 30/kg

Let original price of 1 ltr. Petrol bex.

If price increased by Rs.5 per ltr.

⇒ New price = x + 5 per ltr.

Given, one gets 2 ltr. Petrol for Rs.1320.

⇒ 1320(x + 5) – 1320x = 2x(x + 5)

⇒ 1320x + 6600x – 1320x – 2 x² + 10x = 0

⇒ 2x² + 10x – 6600 = 0

Dividing the equation by 2, we get –

⇒x² + 5x – 3300 = 0

⇒x² + (60 – 55)x – 3300 = 0

⇒x² + 60x – 55x – 3300 = 0

⇒ x(x + 60) – 55(x + 60) = 0

⇒ x = – 60 or x = 55

But price cannot be negative.

⇒ x = Rs.55/ltr.

And increased price = 60/ltr.

let C.P of flowerpot be x.

⇒ Profit = x%

If C.P = 100, then profit = x.

And, if C.P = x, then Profit –

We know, C.P + Profit = S.P

⇒ 100x + x² = 9600

⇒ x² + 100x – 9600 = 0

⇒ x² + (160 – 60)x – 9600 = 0

⇒ x² + 160x – 60x – 9600 = 0

⇒ x(x + 160) – 60(x + 160) = 0

⇒ (x – 60)(x + 160) = 0

⇒ x – 60 = 0 or x + 160 = 0

⇒ x = 60 or x = – 160

But price cannot be negative.

⇒ x(C.P) = 60

And profit obtained = 60%

Let C.P of pen be x

⇒ Loss = x%

If C.P = Rs.100, then loss = Rs. X

And, if C.P = x, then

Also, Loss = C.P – S.P

⇒ x² – 100x + 2400 = 0

⇒ x² – (60 + 40)x + 2400 = 0

⇒ x² – 60x – 40x + 2400 = 0

⇒ x – 60 = 0 or x – 40 = 0

⇒ x = 60 or x = 40

But given C.P is less than 50

⇒ Cost Price(C.P) = Rs.40

Let the length of perpendicular be x.

Now, given that the difference of perpendicular and base = 3

⇒ Base = x + 3

Also, Perimeter = 36

⇒ 36 = x + (x + 3) + Hypotenuse

⇒ Hypotenuse = 36 – x – x – 3

⇒ 33 – 2x

Now, using Pythagoras Theorem,

Hypotenuse² = Base² + Perpendicular²

⇒ (33 – 2x)² = x² + (x + 3)²

⇒ 1089 + 4x² – 132x = x² + x² + 9 + 6x

⇒ 2x² – 1138x + 1080 = 0

Dividing whole equation by 2, we get –

⇒ x² – 69x + 540 = 0

⇒ x² – (60 + 9)x + 540 = 0

⇒ x² – 60x – 9x + 540 = 0

⇒ x(x – 60) – 9(x – 60) = 0

⇒ (x – 60)(x – 9) = 0

⇒ x = 60 or x = 9

∵ Perimeter is given to be 36 cm

⇒ x(Perpendicular) = 9 cm

⇒ Base = x + 3

= 12 cm

Now,

⇒ 54 cm²

According to Pythagoras theorem,

Hypotenuse² = Base² + Perpendicular²

⇒ (x + 6)² = x² + (x + 3)²

⇒ x² + 36 + 12x = x² + x² + 9 + 6x

⇒ x² – 6x – 27 = 0

⇒ x² – (9 – 3)x – 27 = 0

⇒ x² – 9x + 3x – 27 = 0

⇒ x(x – 9) + 3(x – 9) = 0

⇒ (x – 9)(x + 3) = 0

⇒ x = 9 or x = – 3

But side can’t be negative.

⇒ x(Base) = 9 cm.

⇒ Perpendicular = x + 3

= 12 cm

And, hypotenuse = x + 6

= 15 cm

∴ the Perimeter = sum of all 3 sides

= 9 + 12 + 15

= 36 cm

x² — 3x + 2 = 0

⇒ x² —(2 + 1)x + 2 = 0

⇒ x² — 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 1) = 0 or (x – 2) = 0

⇒ x = 1 or x = 2

(B) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

We know,

D = b² – 4ac

Here, a = 5, b = – 6 and c = 1

⇒ D = (– 6)² – 4(5)(1)

⇒ D = 36 – 20

⇒ D = 16

(B) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

∵ 2 is a root

⇒ (2)² – 4(2) + a = 0

⇒ 4 – 8 + a = 0

⇒ a = 4

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

when D = 0, the quadratic equation has equal roots.

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

We’ll put x = 3 in all equation and will see if that satisfies.

For (A), x² — x — 6 = 0

⇒ 9 – 3 – 6 = 0 satisfied.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

∵ 4 is a root –

⇒ 4² + a(4) – 8 = 0

⇒ 16 + 4a – 8 = 0

⇒ a = – 2

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(D) doesn’t match the solution.

∵ 3 is a root

⇒ k(3)² – 7(3) + 3 = 0

⇒ 9k – 21 + 3 = 0

⇒ 9k = 18

⇒ k = 2

(A) doesn’t match the solution.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

We know,

D = b² – 4ac

Here, a = 1, b = – 3 and c = – k

⇒ D = (– 3)² – 4(1)(– k) = 1

⇒ 9 + 4k = 1

⇒ 4k = – 8

⇒ k = – 2

(A) doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.