Question

For what value of k the roots of the quadratic equation kx(x-2√5)+10=0 are equal?
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

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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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Answer 2

$kx(x-2\sqrt{5} )x+10=0$

$kx^2-2\sqrt{5} kx+10=0$

Compare the given equation with the general form of the quadratic equation, which

$ax^2+bx+c=0$

$a=k,b=-2\sqrt{5}k,c=10$

Find discriminant:

$D=b^2-4ac$

$=(-2k)^2-4\times k\times 10=20k^2-40k$

Since roots are real and equal (given), put D=0

$20k^2-40k=0$

$k^2-2k=0$

$k(k-2)=0$

Either, $k=0 or k-2=0$

Hence $k=0$ or $k=2$

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