Question

Find that non-zero value of k, for which the quadratic equation kx² + 1 − 2(k − 1)x + x² = 0 has equal roots. Hence find the roots of the equation.

Answer 1

We have

kx² +1−2(k−1)x+x²=0

This equation can be rearranged as

(k+1)x² −2(k−1)x+1=0

Here, a = k + 1, b = −2(k − 1) and c = 1

∴ D = b² − 4ac

=[−2(k−1)²]−4×(k+1)×1

=4(k−1)²−4(k+1)

=4[(k−1)² −k−1] 

=4[k² +1−2k−k−1]

=4[k²−3k]

=4[k(k−3)]

The given equation will have equal roots, if D = 0

⇒ 4[k(k−3)] = 0

⇒ k = 0 or k − 3 = 0

⇒ k = 3

Putting k = 3 in the given equation, we get

4x²−4x+1=0 

⇒(2x−1)²=0

⇒2x−1=0

${\displaystyle \Rightarrow x=\frac{1}{2}}$

Hence, the roots of the given equation are ${\displaystyle \frac{1}{2}\text{and}\frac{1}{2}}$