Class 10 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 2: Quadratic Equations

Quadratic equations:

x² - 5x + 6 = 0

x² - 9 = 0

The given equation is x² + 5x - 2 = 0

Here, x is the only variable and maximum index of the variable is 2.

Also, a = 1, b = 5, c = -2 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

The given equation is y² = 5y - 10

∴ y²- 5y + 10 = 0

Here, y is the only variable and maximum index of the variable is 2.

Also, a = 1, b = -5, c = 10 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

The given equation is

Multiplying both sides by of the equation by y

∴ y³ + 1 = 2y

∴ y³- 2y + 1 = 0

Here, y is the only variable and maximum index of the variable is not 2.

∴ The given equation is not a quadratic equation.

The given equation is

Multiplying both sides by x

∴ x² + 1 = -2x

∴ x² + 2x+ 1 = 0

Here, x is the only variable and maximum index of the variable is 2.

Also, a = 1, b = 2, c = 1 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

The given equation is (m + 2)(m - 5) = 0

Expanding the above equation, we get

m² + 2m - 5m - 10 = 0

∴ m² - 3m - 10 = 0

Here, m is the only variable and maximum index of the variable is 2.

Also, a = 1, b = -3, c = -10 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

The given equation is m³ + 3m² - 2 = 3m³

∴ m³ - 3m³ + 3m² - 2 = 0

∴ -2m³ + 3m² - 2 = 0

Here, m is the only variable and maximum index of the variable is 3.

∴ The given equation is not a quadratic equation.

The given equation is 2y = 10 - y²

∴ y² + 2y - 10 = 0

This is a quadratic equation.

Comparing the above equation with ay² + by + c = 0, we get

a = 1, b = 2 and c = -10

The given equation is (x - 1)² = 2x + 3

∴ x² - 2x + 1 = 2x + 3

∴ x² - 2x + 1 - 2x - 3 = 0

∴ x² - 4x - 2 = 0

This is a quadratic equation.

Comparing the above equation with ax² + bx + c = 0, we get

a = 1, b = -4 and c = -2

The given equation is x² + 5x = -(3 - x)

∴ x² + 5x + (3 - x) = 0

∴ x² + 4x + 3 = 0

This is a quadratic equation.

Comparing the above equation with ax² + bx + c = 0, we get

a = 1, b = 4 and c = 3

The given equation is 3m² = 2m² - 9

∴ 3m² - 2m² + 9 = 0

∴ m² + 9 = 0

This is a quadratic equation.

Comparing the above equation with am² + bm + c = 0, we get

a = 1, b = 0 and c = 9

The given equation is p(3 + 6p) = -5

∴ 3p + 6p² + 5 = 0

∴ 6p² + 3p + 5 = 0

This is a quadratic equation.

Comparing the above equation with ap² + bp + c = 0, we get

a = 6, b = 3 and c = 5

The given equation is x² - 9 = 13

∴ x² - 9 - 13 = 0

∴ x² - 22 = 0

This is a quadratic equation.

Comparing the above equation with ax² + bx + c = 0, we get

a = 1, b = 0 and c = -22

The given equation is x² + 4x - 5 = 0 … (I)

Substituting x = 1 in L.H.S. of equation (I), we get

L.H.S. = (1)² + 4(1) - 5 = 1 + 4 - 5 = 0

∴ L.H.S. = R.H.S.

∴ x = 1 is the root of the given quadratic equation.

Substituting x = -1 in L.H.S. of equation (I), we get

L.H.S. = (-1)² + 4(-1) - 5 = 1 - 4 - 5 = -8

∴ LH.S. ≠ R.H.S.

∴ x = -1 is not the root of the given quadratic equation.

The given equation is 2m2 - 5m = 0 … (I)

Substituting m = 2 in L.H.S. of equation (I), we get

L.H.S. = 2(2)2 - 5(2) = 2(4) -10 = 8 - 10 = -2

∴ L.H.S. ≠ R.H.S.

∴ m = 2 is not the root of the given quadratic equation.

Putting m = in L.H.S. of equation (I), we get

L.H.S

∴ L.H.S. = R.H.S.

∴ m = is the root of the given quadratic equation.

Given quadratic equation is kx² - 10x + 3 = 0 … (I)

Since, x = 3 is the root of equation (I), we have

k(3)² - 10(3) + 3 = 0

∴ 9k - 30 + 3 = 0

∴ 9k - 27 = 0

∴ 9k = 27

∴ k = 3

is a root of quadratic equation 5m² + 2m + k = 0

∴ Put m =   in the equation.

The given quadratic equation is x² - 15x + 54 = 0

∴ x² - 9x - 6x + 54 = 0

∴ x(x - 9) - 6(x - 9) = 0

∴ (x - 9)(x - 6) = 0

∴ x - 9 = 0 or x - 6 = 0

∴ x = 9 or x = 6

∴ 9 and 6 are the roots of the given quadratic equation.

The given quadratic equation is x² + x - 20 = 0

∴ x² + 5x - 4x - 20 = 0

∴ x² + 5x - 4x - 20 = 0

∴ x(x + 5) - 4(x + 5) = 0

∴ (x + 5)(x - 4) = 0

∴ x + 5 = 0 or x - 4 = 0

∴ x = -5 or x = 4

∴ -5 and 4 are the roots of the given quadratic equation.

The given quadratic equation is 2y² + 27y + 13 = 0

∴ 2y² + 26y + y + 13 = 0

∴ 2y² + 26y + y + 13 = 0

∴ 2y(y + 13) + 1(y + 13) = 0

∴ (y + 13)(2y + 1) = 0

∴ y + 13 = 0 or 2y + 1 = 0

∴ y = -13 or 2y = -1

∴ y = -13 or y =

∴ -13 and are the roots of the given quadratic equation.

The given quadratic equation is 5m² = 22m + 15

∴ 5m² = 22m + 15

∴ 5m² - 25m + 3m - 15 = 0

∴ 5m² - 25m + 3m - 15 = 0

∴ 5m(m - 5) + 3(m - 5) = 0

∴ (m - 5)(5m + 3) = 0

∴ m - 5 = 0 or 5m + 3 = 0

∴ m = 5 or 5m = -3

∴ m = 5 or y =

∴ 5 and are the roots of the given quadratic equation.

The given quadratic equation is 2x² - 2x += 0

Multiplying both sides of the above equation by 2

∴ 4x² - 4x + 1 = 0

∴ 4x² - 2x - 2x + 1 = 0

∴ 4x² - 2x - 2x + 1 = 0

∴ 2x(2x - 1) - (2x - 1) = 0

∴ (2x - 1)(2x - 1) = 0

∴ 2x - 1 = 0 or 2x - 1 = 0

∴ x =  or x =

∴ and  are the roots of the given quadratic equation.

The given quadratic equation is 6x - = 1

Multiplying both sides of the above equation by x

∴ 6x² - 2 = x

∴ 6x² - x - 2 = 0

∴ 6x² - 4x + 3x - 2 = 0

∴ 6x² - 4x + 3x - 2 = 0

∴ 2x(3x - 2) + 1(3x - 2) = 0

∴ (3x - 2)(2x + 1) = 0

∴ 3x - 2 = 0 or 2x + 1 = 0

∴ x =  or x =

∴ and  are the roots of the given quadratic equation.

The given quadratic equation is 3x² - + 2 = 0

∴  

∴ and  are the roots of the given quadratic equation.

The given quadratic equation is 2m(m - 24) = 50

Dividing the above equation by 2

∴ m(m - 24) = 25

∴ m² - 24m = 25

∴ m² - 24m - 25 = 0

∴ m² - 25m + m - 25 = 0

∴ m² - 25m + m - 25 = 0

∴ m(m - 25) + 1(m - 25) = 0

∴ (m - 25)(m + 1) = 0

∴ m - 25 = 0 or m + 1 = 0

∴ m = 25 or m = -1

∴ 25 and -1 are the roots of the given quadratic equation.

The given quadratic equation is 25m² = 9

∴ 25m² - 9 = 0

∴ (5m)² - (3)² = 0

Since, a² - b² = (a - b)(a + b)

∴ (5m - 3)(5m + 3) = 0

∴ 5m - 3 = 0 or 5m + 3 = 0

∴ 5m = 3 or 5m = -3

∴

∴ and  are the roots of the given quadratic equation.

The given quadratic equation is 7m² = 21m

∴ 7m² - 21m = 0

Dividing both sides of the above equation by 7

∴ m² - 3m = 0

∴ m(m - 3) = 0

∴ m = 0 or m - 3 = 0

∴ m = 0 or m = 3

∴ 0 and 3 are the roots of the given quadratic equation.

The given quadratic equation is m² - 11 = 0

Since, a² - b² = (a - b)(a + b)

∴ and are the roots of the given quadratic equation.

If x² + x - 20 = (x + a)²

∴ x² + x + k = x² + 2ax + a²

Comparing the terms, we get

x = 2ax and k = a²

∴ and

Now, x² + x - 20 = 0

Taking square root on both the sides, we get

∴ 4 and -5 are the roots of the given quadratic equation.

If x² + 2x - 5 = (x + a)²

∴ x² + 2x + k = x² + 2ax + a²

Comparing the terms, we get

2x = 2ax and k = a²

∴ a = 1 and k = 1

Now, x² + 2x - 5 = 0

∴ x² + 2x + 1 - 1 - 5 = 0

∴ (x + 1)² - 6 = 0

∴ (x + 1)² = 6

Taking square root on both the sides, we get

∴ and are the roots of the given quadratic equation.

The given quadratic equation is m² - 5m = -3

∴ m² - 5m + 3 = 0

If m² - 5m + k = (m + a)²

∴ m² - 5m + k = m² + 2am + a²

Comparing the terms, we get

-5m = 2am and k = a²

Now, m² - 5m + 3 = 0

Taking square root on both the sides, we get

∴ and are the roots of the given quadratic equation.

The given quadratic equation is 9y² - 12y + 2 = 0

Dividing both sides by 9, we get

If

Comparing the terms, we get

Now,

Taking square root on both the sides, we get

∴ and are the roots of the given quadratic equation.

The given quadratic equation is 2y² + 9y + 10 = 0

Dividing both sides by 2, we get

If

Comparing the terms, we get

Now,

Taking square root on both the sides, we get

∴ -2 and are the roots of the given quadratic equation.

The given quadratic equation is 5x² = 4x + 7

∴ 5x² - 4x - 7 = 0

Dividing both sides by 5, we get

If

Comparing the terms, we get

Now,

Taking square root on both the sides, we get

∴ and are the roots of the given quadratic equation.

The above equation is x² - 7x + 5 = 0 … (I)

Comparing above equation (I) with ax² + bx + c = 0, we get

a = 1, b = -7 and c = 5

The above equation is 2m² = 5m - 5

∴ 2m² - 5m + 5 = 0 … (I)

Comparing above equation (I) with ax² + bx + c = 0, we get

a = 2, b = -5 and c = 5

The above equation is y² = 7y

∴ y² - 7y = 0 … (I)

Comparing above equation (I) with ax² + bx + c = 0, we get

a = 1, b = -7 and c = 0

The above equation is x² + 6x + 5 = 0 … (I)

Comparing above equation (I) with ax² + bx + c = 0, we get

a = 1, b = 6 and c = 5

∴ b² - 4ac = (6)² - (4 × 1 × 5)

= 36 - 20

= 16

∴ x = -1 or x = -5

∴ The roots of the given quadratic equation are -1 and -5.

The above equation is x² - 3x - 2 = 0 … (I)

Comparing above equation (I) with ax² + bx + c = 0, we get

a = 1, b = -3 and c = -2

∴ b² - 4ac = (-3)² - 4 × 1 × (-2)

= 9 + 8

= 17

∴ The roots of the given quadratic equation are and

The above equation is 3m² + 2m - 7 = 0 … (I)

Comparing above equation (I) with am² + bm + c = 0, we get

a = 3, b = 2 and c = -7

∴ b² - 4ac = (2)² - 4 × 3 × (-7)

= 4 + 84

= 88

∴ The roots of the given quadratic equation are and

The above equation is 5m² - 4m - 2 = 0 … (I)

Comparing above equation (I) with am² + bm + c = 0, we get

a = 5, b = -4 and c = -2

∴ b² - 4ac = (-4)² - 4 × 5 × (-2)

= 16 + 40

= 56

∴ The roots of the given quadratic equation are and

The above equation is y² + = 2

∴ 3y² + y = 6

∴ 3y² + y - 6 = 0 … (I)

Comparing above equation (I) with ay² + by + c = 0, we get

a = 3, b = 1 and c = -6

∴ b² - 4ac = (1)² - 4 × 3 × (-6)

= 1 + 72

= 73

∴ The roots of the given quadratic equation are and

The above equation is 5x² + 13x + 8 = 0 … (I)

Comparing above equation (I) with ay² + by + c = 0, we get

a = 5, b = 13 and c = 8

∴ b² - 4ac = (13)² - 4 × 5 × (8)

= 169 - 160

= 9

∴ The roots of the given quadratic equation are -1 and

Given equation is

Comparing the above equation with ax2 + bx + cx = 0, we get

∴ The roots of the given quadratic equation are

(1)

(2)

(3)

(1)

The above equation is x² + 7x - 1 = 0 … (I)

Comparing above equation (I) with ax² + bx + c = 0, we get

a = 1, b = 7 and c = -1

∴ b² - 4ac = (7)² - 4 × 1 × (-1)

= 49 + 4

= 53

∴ Discriminant = 53

(2)

The above equation is 2y² - 5y + 10 = 0 … (I)

Comparing above equation (I) with ay² + by + c = 0, we get

a = 2, b = -5 and c = 10

∴ b² - 4ac = (-5)² - 4 × 2 × 10

= 25 - 80

= -55

∴ Discriminant = -55

(3)

The above equation is x² + 4x + = 0 … (I)

Comparing above equation (I) with ax² + bx + c = 0, we get

a =, b = 4 and c =

∴ b² - 4ac = (4)² - 4 × ×

= 16 - 16

= 0

∴ Discriminant = 0

(1)

The above equation is x² - 4x + 4 = 0 … (I)

Comparing above equation (I) with ax² + bx + c = 0, we get

a = 1, b = -4 and c = 4

∴ b² - 4ac = (-4)² - 4 × 1 × 4

= 16 - 16

= 0

Since, ∆ = 0

∴ Roots of the given quadratic equation are real and equal.

(2)

The above equation is 2y² - 7y + 2 = 0 … (I)

Comparing above equation (I) with ay² + by + c = 0, we get

a = 2, b = -7 and c = 2

∴ b² - 4ac = (-7)² - 4 × 2 × 2

= 49 - 16

= 33

Since, ∆ > 0

∴ Roots of the given quadratic equation are real and distinct.

(3)

The above equation is m² + 2m + 9 = 0 … (I)

Comparing above equation (I) with am² + bm + c = 0, we get

a = 1, b = 2 and c = 9

∴ b² - 4ac = 2² - 4 × 1 × 9

= 4 - 36

= -32

Since, ∆ < 0

∴ Roots of the given quadratic equation are not real.

(1)

Let α = 0 and β = 4

∴ α + β = 0 + 4 = 4 and α×β = 0 × 4 = 0

The quadratic equation is

x² - (α + β) x + α×β = 0

∴ x² - 4x + 0 = 0

∴ x² - 4x = 0

(2)

Let α = 3 and β = -10

∴ α + β = 3 + (-10) = -7 and α×β = 3 × (-10) = -30

The quadratic equation is

x² - (α + β) x + α×β = 0

∴ x² - (-7) x + (-30) = 0

∴ x² +7x - 30 = 0

(3)

Let

∴  

The quadratic equation is

x² - (α + β) x + α×β = 0

(4)

Let

The quadratic equation is

x² - (α + β) x + α×β = 0

∴ x² - 4x + (-1) = 0

∴ x² - 4x - 1 = 0

The equation is

x² - 4kx + k + 3 = 0 … (I)

Comparing equation (I) with the equation ax² + bx + c = 0, we get

a = 1, b = -4k and c = k + 3 … (II)

Let α and β be the roots of the quadratic equation.

As per the condition given in the question, we have

α+β = 2 αβ

… From equation (II)

∴ 4k - 2k = 6

∴ 2k = 6

∴ k = 3

Given quadratic equation is y² - 2y - 7 = 0 … (I)

Comparing equation (I) with the equation ay² + by + c = 0, we get

a = 1, b = -2 and c = -7

Since, α and β be the roots of equation (I)

(1)

Now,

(2)

Also,

(1)

The quadratic equation is 3y² + ky + 12 = 0

Comparing the above equation with the equation ay² + by + c = 0, we get

a = 3, b = k and c = 12

Since, the roots are real and equal.

∴ Discriminant (∆) = 0

∴ b² - 4ac = 0

∴ k² - 4(3)(12) = 0

∴ k² - 12² = 0

Since, a² - b² = (a - b)(a + b)

∴ (k - 12)(k + 12) = 0

∴ k - 12 = 0 or k + 12 = 0

∴ k = 12 or k = -12

(2)

The quadratic equation is kx (x - 2) + 6 = 0

∴ kx² - 2kx + 6 = 0

Comparing the above equation with the equation ax² + bx + c = 0, we get

a = k, b = -2k and c = 6

Since, the roots are real and equal.

∴ Discriminant (∆) = 0

∴ b² - 4ac = 0

∴ (-2k)² - 4(k)(6) = 0

∴ 4k² - 24k = 0

∴ 4k(k - 6) = 0

∴ 4k = 0 or k - 6 = 0

∴ k = 0 or k = 6

But k = 0 is not possible as the quadratic coefficient becomes zero.

∴ k = 6

Let the present age of Pragati be x years.

2 years ago, age of Pragati = (x - 2) years

After 3 years, age of Pragati = (x + 3) years

As per the condition given in the question,

(x - 2)(x + 3) = 84

∴ x(x + 3) - 2(x + 3) = 84

∴ x² + 3x - 2x - 6 = 84

∴ x² + x - 6 - 84 = 0

∴ x² + x - 90 = 0

∴ x² + 10x - 9x - 90 = 0

∴ x(x + 10) - 9(x + 10) = 0

∴ (x + 10)(x - 9) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x + 10 = 0 or x - 9 = 0

∴ x = -10 or x = 9

Since the age can't be negative, we get

x = 9

Hence, Pragati's present age is 9 years.

Let the first even natural number be x.

∴ The next consecutive even natural number will be (x + 2).

From the given condition, we have

x² + (x + 2)² = 244

∴ x² + x² + 4x + 4 = 244

∴ 2x² + 4x + 4 - 244 = 0

∴ 2x² + 4x - 240 = 0

Dividing both sides of above equation by 2

∴ x² + 2x - 120 = 0

∴ x² + 12x - 10x - 120 = 0

∴ x(x + 12) - 10(x + 12) = 0

∴ (x + 12)(x - 10) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x + 12 = 0 or x - 10 = 0

∴ x = -12 or x = 10

But, natural number cannot be negative.

∴ x = 10

And, x + 2 = 10 + 2 = 12

∴ The two consecutive even natural numbers are 10 and 12.

Number of trees in a column is x.

∴ Number of trees in a row = x + 5

∴ Total number of trees = x × (x + 5)

From the given condition, we have

x(x + 5) = 150

∴ x² + 5x = 150

∴ x² + 5x - 150 = 0

∴ x² + 15x - 10x - 150 = 0

∴ x(x+ 15) - 10(x + 15) = 0

∴ (x + 15)(x - 10) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x + 15 = 0 or x - 10 = 0

∴ x = -15 or x = 10

Since, the number of trees cannot be negative.

∴ x = 10

∴ Number of trees in a column is 10.

∴ Number of trees in a row = x + 5 = 10 + 5 = 15

∴ Number of trees in a row is 15.

Let the present age of Kishor be x years.

∴ Present age of Vivek = (x + 5) years

From the given condition, we have

∴ x²- 7x - 30 = 0

∴ x²- 10x + 3x - 30 = 0

∴ x(x - 10) + 3(x - 10) = 0

∴ (x - 10)(x + 3) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x - 10 = 0 or x + 3 = 0

∴ x = 10 or x = -3

Since, the age cannot be negative.

∴ x = 10

And, x + 5 = 10 + 5 = 15

∴ Present ages of Kishor and Vivek are 10 years and 15 years respectively.

Let Suyash score x marks in the first test.

∴ Score in the second test = x + 10

From the given condition, we have

5(x + 10) = x²

∴ 5x + 50 = x²

∴ x²- 5x - 50 = 0

∴ x²- 10x + 5x - 50 = 0

∴ x(x - 10) + 5(x - 10) = 0

∴ (x - 10)(x + 5) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x - 10 = 0 or x + 5 = 0

∴ x = 10 or x = - 5

Since, the score cannot be negative.

∴ x = 10

∴ The score of Suyash in the first test is 10 marks.

Let Mr. Kasam make x number of pots on daily basis.

Production cost of each pot = Rs. (10x + 40)

From the given condition, we have

x(10x + 40) = 600

∴ 10x² + 40x = 600

∴ 10x² + 40x - 600 = 0

Dividing both sides of above equation by 10, we get

x² + 4x - 60 = 0

∴ x² + 10x - 6x - 60 = 0

∴ x(x + 10) - 6(x + 10) = 0

∴ (x + 10)(x - 6) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x + 10 = 0 or x - 6 = 0

∴ x = -10 or x = 6

Since, the number of pots cannot be negative.

∴ x = 6

∴ Production cost of each pot = (10x + 40)

= Rs [(10×6)+ 40]

= Rs (60 + 40)

= Rs 100

Production cost of one pot is Rs 100 and the number of pots Mr. Kasam makes per day is 6.

Let the speed of water current be x km/hr.

Speed of boat in still water is given as 12 km/hr. (x < 12)

In upstream, speed of the water current decreases the speed of the boat.

Whereas, in downstream, speed of the water current increases the speed of the boat.

∴ speed of the boat in upstream = (12 - x) km/hr

and speed of the boat in downstream = (12 + x) km/hr

As,

∴ Time required to cover 36 km upstream hrs

Time required to cover 36 km downstream hrs

From the given condition, we have

∴ 216 = 288 - 2x²

∴ 2x² + 216 - 288 = 0

∴ 2x² - 72 = 0

Dividing both sides of above equation by 2, we get

x² - 36 = 0

∴ x² = 36

Taking square root on both the sides

∴ x = ±6

Since, the speed can't be negative.

∴ x = 6

∴ The speed of water current is 6 km/hr.

Let Nishu take x days to complete the work alone.

∴ Total work done by Nishu in 1 day

Also, Pintu takes (x + 6) days to complete the work alone.

∴ Total work done by Pintu in 1 day

∴ Total work done by both in 1 day

It is given that both take 4 days to complete the work together.

∴ Total work done by both in 1 day

From the given condition, we have

∴ 4(2x + 6) = x(x + 6)

∴ 8x + 24 = x² + 6x

∴ x² + 6x - 8x - 24 = 0

∴ x²- 2x - 24 = 0

∴ x²- 6x + 4x - 24 = 0

∴ x(x - 6)+ 4(x - 6) = 0

∴ (x - 6) (x + 4) = 0

∴ x - 6 = 0 or x + 4 = 0

∴ x = 6 or x = -4

But, number of days cannot be negative,

∴ x = 6 and x + 6 = 6 + 6 = 12

∴ Number of days taken by Nishu and Pintu to complete the work alone is 6 days and 12 days respectively.

Let the natural number be x.

∴ Divisor = x, Quotient = 5x + 6

Also, Dividend = 460 and Remainder = 1

Since, Dividend = Divisor × Quotient + Remainder

∴ 460 = x × (5x + 6) + 1

∴ 460 = 5x² + 6x + 1

∴ 5x² + 6x + 1 - 460 = 0

∴ 5x² + 6x - 459 = 0

∴ 5x² - 45x + 51x - 459 = 0

∴ 5x(x - 9) + 51(x - 9) = 0

∴ (x - 9)(5x + 51) = 0

∴ x - 9 = 0 or 5x + 51 = 0

∴ x = 9 or

Since, the natural number cannot be negative,

∴ x = 9

∴ Quotient = 5x + 6 = 5(9) + 6 = 45 + 6 = 51

∴ Quotient is 51 and divisor is 9.

Given: ⎕ABCD is a trapezium and AB || CD

Now, Area of trapezium  

But length is never negative.

AB = 7 cm, CD = 15 cm, AD = BC = 5 cm

The equation x(x + 5) = 2

i.e. x² + 5x = 2

It is a quadratic equation as the degree is 2.

The equation x² + 4x = 11 + x²

i.e. 4x - 11 = 0

This is not quadratic as the degree of this equation is 1.

The given equation is x² + kx + k = 0

Comparing above equation with ax² + bx + c = 0, we get

a = 1, b = k, c = k

Since, the roots are real and equal.

∴ ∆ = 0

∴ b² - 4ac = 0

∴ k² - 4×1×k = 0

∴ k² - 4k = 0

∴ k(k - 4) = 0

∴ k = 0 or k - 4 = 0

∴ k = 0 or k = 4

Given quadratic equation is √2x² - 5x + √2 = 0

Comparing above equation with ax² + bx + c = 0, we get

Discriminant (∆) = b² - 4ac

∴ Discriminant = 17

Let α = 3 and β = 5

The quadratic equation with the roots α and β is given by

x² - (α + β)x + α×β = 0

∴ x² - (3+5)x + 3×5 = 0

∴ x² - 8x + 15 = 0

∴ The required quadratic equation is x² - 8x + 15 = 0.

For a quadratic equation ax² + bx + c = 0

Sum of the roots

In quadratic equation, 3x² + 15x +3 = 0

∴ Sum of roots = -5

∴ The quadratic equation whose sum of roots is -5 is 3x² + 15x +3 = 0.

The given quadratic equation is

Dividing the above equation by we get

m² - m + 1 = 0

Comparing above equation with am² + bm + c = 0, we get

a = 1, b = -1, c = 1

Now, discriminant (∆) = b² - 4ac

∴ ∆ = (-1)² - 4(1)(1) = 1 - 4 = -3

Since ∆ < 0

So, the roots are not real.

Given equation is x² + mx - 5 = 0 … (I)

As 2 is one of the roots of equation (I)

∴ 2² + m × 2 - 5 = 0

∴ 4 + 2m - 5 = 0

∴ 2m - 1 = 0

(1)

The equation is x² + 2x + 11 = 0

Here, x is the only variable and maximum index of the variable is 2.

Also, a = 1, b = 2, c = 11 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

(2)

The equation is x² - 2x + 5 = x²

i.e. -2x + 5 = 0

Here, x is the only variable and maximum index of the variable is 1.

∴ The given equation is not a quadratic equation.

(3)

The equation is (x + 2)² = 2x²

i.e. x² + 4x + 4 = 2x²

i.e. x² - 4x - 4 = 0

Here, x is the only variable and maximum index of the variable is 2.

Also, a = 1, b = -4, c = -4 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

(1)

The equation is 2y² - y + 2 = 0

Comparing above equation with ay² + by +c = 0, we get

a = 2, b = -1, c = 2

Discriminant (∆) = b² - 4ac

∴ ∆ = (-1)² - 4(2)(2)

= 1 - 16

= -15

(2)

The equation is 5m² - m = 0

Comparing above equation with am² + bm +c = 0, we get

a = 5, b = -1, c = 0

Discriminant (∆) = b² - 4ac

∴ ∆ = (-1)² - 4(5)(0)

= 1 - 0

= 1

(3)

The equation is

Comparing above equation with ax² + bx +c = 0, we get

Discriminant (∆) = b² - 4ac

= 1 + 20

= 21

As -2 is one of the roots of the equation 2x² + kx - 2 = 0.

∴ Putting x = -2 in the given equation, we get

2(-2)² + k(-2) - 2 = 0

∴ 8 - 2k - 2 = 0

∴ 6 - 2k = 0

∴ 2k = 6

∴ k =

∴ k = 3

(1)

Let α = 10 and β = -10

∴ α + β = 10 - 10 = 0 and α×β = 10 × (-10) = -100

The quadratic equation is

x² - (α + β) x + α×β = 0

∴ x² - 0×x + (-100) = 0

∴ x² - 100 = 0

(2)

Let

∴

The quadratic equation is

x² - (α + β) x + α×β = 0

∴ x² - 2x + (-44) = 0

∴ x² - 2x - 44 = 0

(3)

Let α = 0 and β = 7

∴ α + β = 0 + 7 = 7 and αβ = 0×7 = 0

The quadratic equation is

x² - (α + β) x + α×β = 0

∴ x² - 7x + 0 = 0

∴ x² - 7x = 0

(1)

The equation is 3x² - 5x + 7 = 0

Comparing above equation with ax² + bx + c = 0, we get

a = 3, b = -5, c = 7

Discriminant (∆) = b² - 4ac

∴ ∆ = (-5)² - 4(3)(7)

= 25 - 84

= -59

Since, ∆ < 0

∴ The roots are not real.

(2)

The equation is √3x² + √2x - 2√3 = 0

Comparing above equation with ax² + bx + c = 0, we get

Discriminant (∆) = b² - 4ac

Since, ∆ > 0

∴ The roots are real and unequal.

(3)

The equation is m² - 2m + 1 = 0

Comparing above equation with am² + bm + c = 0, we get

a = 1, b = -2, c = 1

Discriminant (∆) = b² - 4ac

∴ ∆ = (-2)² - 4(1)(1)

= 4 - 4

= 0

Since, ∆ = 0

∴ The roots are real and equal.

Given equation is

Cross multiplying both the sides, we get

x² = x + 5

∴ x² - x - 5 = 0 … (I)

Comparing equation (I) with ax² + bx + c = 0, we get

a = 1, b = -1, c = -5

∴ b² - 4ac = (-1)² - 4(1)(-5)

= 1 + 20

= 21

Now,

∴ The roots of the given quadratic equation are

Given equation is

Multiplying both sides by 10, we get

10x² - 3x - 1 = 0

∴ 10x² - 5x + 2x - 1 = 0

∴ 5x(2x - 1) + 1(2x - 1) = 0

∴ (2x - 1)(5x + 1) = 0

∴ 2x - 1 = 0 or 5x + 1 = 0

∴ The roots of the given quadratic equation are

Given equation is (2x + 3)² = 25

∴ 4x² + 12x + 9 = 25

∴ 4x² + 12x + 9 - 25 = 0

∴ 4x² + 12x - 16 = 0

Dividing the above equation by 4, we get

∴ x² + 3x - 4 = 0

∴ x² + 4x - x - 4 = 0

∴ x(x + 4) - 1(x + 4) = 0

∴ (x + 4)(x - 1) = 0

∴ x + 4 = 0 or x - 1 = 0

∴ x = -4 or x = 1

∴ The roots of the given quadratic equation are -4 and 1.

Given equation is m² + 5m + 5 = 0 … (I)

Comparing equation (I) with am² + bm + c = 0, we get

a = 1, b = 5, c = 5

∴ b² - 4ac = (5)² - 4(1)(5)

= 25 - 20

= 5

Now,

∴ The roots of the given quadratic equation are

Given equation is 5m² + 2m + 1 = 0

Comparing above equation with am2 + bm + c = 0, we get

a = 5, b = 2, c = 1

∴ b² - 4ac = (2)² - 4(5)(1)

= 4 - 20

= -16

Since, b² - 4ac < 0

∴ The roots of the given equation are not real.

Given equation is x² - 4x - 3 = 0

Comparing above equation with ax² + bx + c = 0, we get

a = 1, b = -4, c = -3

∴ b² - 4ac = (-4)² - 4(1)(-3)

= 16 + 12

= 28

Now,

∴ The roots of the given quadratic equation are

Given quadratic equation is (m - 12)x² + 2(m - 12)x + 2 = 0

Comparing the above equation with ax² + bx + c = 0, we get

a = m - 12, b = 2(m - 12), c = 2

Now,

∆ = b²- 4ac

= [2(m - 12)]² - 4 × (m - 12) × 2

= 4(m - 12)² - 8(m - 12)

= 4(m - 12) (m - 12 - 2)

∴ ∆ = 4(m - 12) (m - 14)

Since, the roots are real and equal.

∴ ∆ = 0

∴ 4(m - 12) (m - 14) = 0

∴ (m - 12) (m - 14) = 0

∴ m - 12 = 0 or m - 14 = 0

∴ m = 12 or m = 14

If m = 12, then quadratic coefficient becomes zero.

∴ m ≠ 12

∴ m = 14

Let α and β be the roots of the quadratic equation.

From the first condition, we have

α + β = 5 … (I)

From the second condition, we have

α³ + β³ = 35 … (II)

Now, (α + β)³ = α³ + 3α²β + 3αβ² + β³

∴ (α + β)³ = α³ + β³ + 3αβ(α + β)

Using equations (I) and (II)

∴ (5)³ = 35 + 3αβ(5)

∴ 125 = 35 + 15αβ

∴ 125 - 35 = 15αβ

∴ 15αβ = 90

∴ αβ = 6

∴ The required quadratic equation is

x² - (α + β)x + αβ = 0

i.e. x²- 5x + 6 = 0

Given quadratic equation is

2x² + 2(p + q)x + p² + q² = 0 … (I)

Comparing the above equation with ax² + bx + c = 0, we get

a = 2, b = 2(p + q), c = p² + q²

Let α and β be the roots of equation (I)

… (II)

And,

Now,

As per the given condition, roots of the required quadratic equation are

(α + β)² and (α - β)²

Now,

sum of the roots = (α + β)² + (α - β)²

= [-(p + q)]² + (-(p - q)²)

= (p + q)² - (p - q)²

= p² + q² + 2pq - (p² + q² - 2pq)

= p² + q² + 2pq - p² - q² + 2pq

= 4pq

Product of roots = (α + β)² × (α - β)²

= [-(p + q)]² [-(p - q)²]

= -(p + q)² (p - q)²

= -[(p + q)(p - q)]²

= -(p² - q²)²

∴ The required quadratic equation is

x² - [(α + β)² + (α - β)²]x + [(α + β)² × (α - β)²] = 0

i.e. x² - (4pq)x - (p² - q²)² = 0

Let the amount Sagar possesses be Rs x.

∴ The amount Mukund possesses = Rs (x + 50)

As per the condition, we have

x(x + 50)= 15000

∴ x² + 50x - 15000 = 0

∴ x² + 150x - 100x - 15000 = 0

∴ x(x + 150) - 100(x + 150) = 0

∴ (x + 150)(x - 100) = 0

∴ x + 150 = 0 or x - 100 = 0

∴ x = -150 or x = 100

Since, amount cannot be negative.

∴ x = 100 and x + 50 = 100 + 50 = 150

∴ The amount possessed by Sagar and Mukund are Rs. 100 and Rs. 150 respectively.

Let the two numbers be x and y (x > y).

As per the given condition, we have

x² - y² = 120 … (I)

y² = 2x … (II)

Substituting y² = 2x in equation (i), we get

x² - 2x = 120

∴ x²- 2x - 120 = 0

∴ x²- 12x + 10x - 120 = 0

∴ x(x - 12) + 10(x - 12) = 0

∴ (x - 12)(x + 10) = 0

∴ x - 12 = 0 or x + 10 = 0

∴ x = 12 or x = -10

When x ≠ -10,

y² = 2x = 2(-10) = -20 … [Since, the square of number cannot be negative]

∴ x = 12

Smaller number = y² =2x

∴ y² = 2 × 12

∴ y² = 24

∴ y = ± √24 …[Taking square root of both sides]

∴ The smaller number is √24 and greater number is 12 or the smaller number is -√24 and greater number is 12.

Let the number of students be x.

Total number of oranges = 540 … (Given)

∴ The number of oranges each student gets

When there are 30 more students, we have

Total number of students = (x + 30)

and total number of oranges each student gets  

As per the given condition, we have

∴ 540×30 = 3x² + 90x

∴ 3x² + 90x - 540×30 = 0

Dividing both sides by 3, we get

x² + 30x - 540×10 = 0

∴ x² + 30x - 5400 = 0

∴ x² + 90x - 60x - 5400 = 0

∴ x(x + 90) - 60(x + 90) = 0

∴ (x + 90)(x - 60) = 0

∴ x + 90 = 0 or x - 60 = 0

∴ x = -90 or x = 60

Since the number of students can't be negative.

∴ x = 60

∴ The total number of students is 60.

Let the breadth of the rectangular farm be x m.

∴ Length of rectangular farm = (2x + 10) m

Now, Area of rectangular farm = Length × Breadth

= (2x + 10) × x

= (2x² + 10x) sq. m

Now, side of square shaped pond m

∴ Area of square shaped pond = (side)²

According to the given condition, we have

Area of rectangular farm = 20 × Area of pond

Dividing both sides of the above equation by 2, we get

∴ x = 0 or x - 45 = 0

∴ x = 0 or x = 45

Since the breadth of the farm can't be zero.

∴ x = 45

Length of the rectangular farm = 2x + 10

= 2×45 + 10

= 90 + 10

= 100 m

Side of the pond m

∴ Length and breadth of the farm and the side of pond are 100 m, 45 m and 15 m respectively.

Let the larger tap take x hours to fill the tank completely.

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

Also, the smaller tap takes (x + 3) hours to fill the tank completely.

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

∴ Part of tank filled by both the taps in 1 hour

But, the tank gets filled in 2 hours by both the taps.

∴ Part of tank filled by both the taps in 1 hour = ½

As per the given condition, we have

∴ 2(2x + 3) = x(x + 3)

∴ 4x + 6 = x² + 3x

∴ x² + 3x - 4x - 6 = 0

∴ x²- x - 6 = 0

∴ x²- 3x + 2x - 6 = 0

∴ x(x - 3) + 2(x - 3) = 0

∴ (x - 3)(x + 2) = 0

∴ x - 3 = 0 or x + 2 = 0

∴ x = 3 or x = -2

But as we know that time cannot be negative.

∴ x = 3 and x + 3 = 3 + 3 = 6

∴ The larger tap takes 3 hours and the smaller tap takes 6 hours to fill the tank completely.