Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

## Practice Set 2.5 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.

Fill in the gaps and complete.

Answer:

Question 2.

Find the value of discriminant.

i. x^{2} + 7x – 1 = 0

ii. 2y^{2} – 5y + 10 = 0

iii. √2 x^{2} + 4x + 2√2 = 0

Solution:

i. x^{2} +7 x – 1 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

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

∴ b^{2}– 4ac = (7)^{2} – 4 × 1 × (-1)

= 49 + 4

∴ b^{2} – 4ac = 53

ii. 2y^{2} – 5y + 10 = 0

Comparing the above equation with

ay^{2} + by + c = 0, we get

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

∴ b^{2 }– 4ac = (-5)2 -4 × 2 × 10

= 25 – 80

∴ b^{2} – 4ac = -55

iii. √2 x^{2} + 4x + 2√2 = 0

Comparing the above equation with

ax + bx + c = 0, we get

a = √2,b = 4, c = 2√2

∴ b^{2} – 4ac = (4)2 – 4 × √2 × 2√2

= 16 – 16

∴ b^{2} – 4ac =0

Question 3.

Determine the nature of roots of the following quadratic equations.

i. x^{2} – 4x + 4 = 0

ii. 2y^{2} – 7y + 2 = 0

iii. m^{2} + 2m + 9 = 0

Solution:

i. x^{2} – 4x + 4= 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

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

∴ ∆ = b^{2} – 4ac

= (- 4)^{2} – 4 × 1 × 4

= 16 – 16

∴ ∆ = 0

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

ii. 2y^{2} – 7y + 2 = 0

Comparing the above equation with

ay^{2} + by + c = 0, we get

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

∴ ∆ = b^{2} – 4ac

= (- 7)^{2} – 4 × 2 × 2

= 49 – 16

∴ ∆ = 33

∴ ∆ > 0

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

iii. m^{2} + 2m + 9 = 0

Comparing the above equation with

am^{2} + bm + c = 0, we get

a = 1,b = 2, c = 9

∴ ∆ = b^{2} – 4ac

= (2)^{2} – 4 × 1 × 9

= 4 – 36

∴ ∆ = -32

∴ ∆ < 0

∴ Roots of the given quadratic equation are not real.

Question 4.

Form the quadratic equation from the roots given below.

i. 0 and 4

ii. 3 and -10

iii. \(\frac { 1 }{ 2 } \) , \(\frac { 1 }{ 2 } \)

iv. 2 – √5, 2 + √5

Solution:

i. Let a = 0 and β = 4

∴ α + β = 0 + 4 = 4

and α × β = 0 × 4 = 0

∴ The required quadratic equation is

x^{2} – (α + β) x + αβ = 0

∴ x^{2} – 4x + 0 = 0

∴ x^{2} – 4x = 0

ii. Let α = 3 and β = -10

∴ α + β = 3 – 10 = -7

and α × β = 3 × -10 = -30

∴ The required quadratic equation is

x^{2} – (α + β)x + αβ = 0

∴ x^{2} – (-7) x + (-30) = 0

∴ The required quadratic equation is

x^{2} – (α + β)x + αβ = 0

∴ x^{2} – 4x – 1 = 0

Question 5.

Sum of the roots of a quadratic equation is double their product. Find k if equation is x^{2} – 4kx + k + 3 = 0.

Solution:

x^{2} – 4kx + k + 3 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

a = 1, b = – 4k, c = k + 3

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

Then, α + β = \(\frac { -b }{ a } \) and αβ = \(\frac { c }{ a } \)

According to the given condition,

Question 6.

α, β are roots of y^{2} – 2y – 7 = 0 find,

i. α^{2} + β^{2}

ii. α^{3} + β^{3}

Solution:

y^{2} – 2y – 7 = 0

Comparing the above equation with

ay^{2} + by + c = 0, we get

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

Question 7.

The roots of each of the following quadratic equations are real and equal, find k.

i. 3y^{2} + ky + 12 = 0

ii. kx (x-2) + 6 = 0

Solution:

i. 3y^{2} + kg + 12 = 0

Comparing the above equation with

ay^{2} + by + c = 0, we get

a = 3, b = k, c = 12

∴ ∆ = b^{2} – 4ac

= (k)^{2} – 4 × 3 × 12

= k^{2} – 144 = k^{2} – (12)^{2}

∴ ∆ = (k + 12) (k – 12) …[∵ a^{2} – b^{2} = (a + b) (a – b)]

Since, the roots are real and equal.

∴ ∆ = 0

∴ (k + 12) (k – 12) = 0

∴ k + 12 = 0 or k – 12 = 0

∴ k = -12 or k = 12

ii. kx (x – 2) + 6 = 0

∴ kx^{2} – 2kx + 6 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

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

∴ ∆ = b^{2} – 4ac

= (-2k)^{2} – 4 × k × 6

= 4k^{2} – 24k

∴ ∆ = 4k (k – 6)

Since, the roots are real and equal.

∴ ∆ = 0

∴ 4k (k – 6) = 0

∴ k(k – 6) = 0

∴ k = 0 or k – 6 = 0

But, if k = 0 then quadratic coefficient becomes zero.

∴ k ≠ 0

∴ k = 6

Question 1.

Fill in the blanks. (Textbook pg. no. 44)

Solution:

Question 2.

Determine nature of roots of the quadratic equation: x^{2} + 2x – 9 = 0 (Textbook pg. no. 45)

Solution:

∴ The roots of the given equation are real and unequal.

Question 3.

Fill in the empty boxes properly. (Textbook pg. no. 46)

Solution:

10x^{2} + 10x + 1 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

a = 10, b = 10, c = 1

Question 4.

Write the quadratic equation if addition of the roots is 10 and product of the roots is 9. (Textbook pg, no. 48)

Answer:

Question 5.

What will be the quadratic equation if α = 2, β = 5. (Textbook pg. no, 48)

Solution: