Chapter 2 Quadratic Equations Practice Set 2.5 Solutions

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

ii. 2y²– 5y + 10 = 0
Comparing the above equation with
ay²+ by + c = 0, we get
a = 2, b = -5, c = 10
∴ b²– 4ac = (-5)2 -4 × 2 × 10
= 25 – 80
∴ b²– 4ac = -55

iii. √2 x²+ 4x + 2√2 = 0
Comparing the above equation with
ax + bx + c = 0, we get
a = √2,b = 4, c = 2√2
∴ b²– 4ac = (4)2 – 4 × √2 × 2√2
= 16 – 16
∴ b²– 4ac =0

ii. 2y²– 7y + 2 = 0
Comparing the above equation with
ay²+ by + c = 0, we get
a = 2, b = -7, c = 2
∴ ∆ = b²– 4ac
= (- 7)²– 4 × 2 × 2
= 49 – 16
∴ ∆ = 33
∴ ∆ > 0
∴ Roots of the given quadratic equation are real and unequal.

iii. m²+ 2m + 9 = 0
Comparing the above equation with
am²+ bm + c = 0, we get
a = 1,b = 2, c = 9
∴ ∆ = b²– 4ac
= (2)²– 4 × 1 × 9
= 4 – 36
∴ ∆ = -32
∴ ∆ < 0
∴ Roots of the given quadratic equation are not real.

ii. Let α = 3 and β = -10
∴ α + β = 3 – 10 = -7
and α × β = 3 × -10 = -30
∴ The required quadratic equation is
x²– (α + β)x + αβ = 0
∴ x²– (-7) x + (-30) = 0

∴ The required quadratic equation is
x²– (α + β)x + αβ = 0
∴ x²– 4x – 1 = 0

ii. kx (x – 2) + 6 = 0
∴ kx²– 2kx + 6 = 0
Comparing the above equation with
ax²+ bx + c = 0, we get
a = k, b = -2k, c = 6
∴ ∆ = b²– 4ac
= (-2k)²– 4 × k × 6
= 4k²– 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