Question

for what value of k the root of quadratic equation kx bracket open x minus 2 root 5 +10 = 0 are equal

Answer 1

Kx(x--2 5^1/2)+10=0
Kx^2--2 5^1/2 Kx+10=0
here,
let a=k; b= --2 5^1/2 k; c=10
A/Q
b^2--4ac=0
(--2 5^1/2 k)^2 -- 4 x k x 10=0
4x5 k^2 -- 40k=0
20k^2--40k=0
k^2--2k=0
k(k--2)=0
k=0; k=2

Answer 2

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kx(x – 2√5 ) + 10 = 0

⇒ kx^2 – 2√5 kx + 10 = 0

Here, a = k, b = – 2√5 k, c = 10

Given roots are equal,

D = b^2 – 4ac = 0

⇒ (–2√5 k)^2 – 4 × k × 10 = 0

⇒ 20k^2 – 40k = 0

⇒ 20k(k – 2) = 0

⇒ k(k – 2) = 0

⇒ k ≠ 0

⇒ k = 2

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