Question

Find the value of k so that following quadritic equation has equal roots : 2x^2- (k-2)x+1=0

Answer 1

Roots are equal, so discriminant is zero
${(k - 2)}^{2} - 4 \times 2 \times 1 = 0 \\ {(k - 2)}^{2} = 8 \\ k - 2 = \sqrt{8} = + \: \: or \: \: - 2 \sqrt{2} \\ k = 2 + 2 \sqrt{2} \: \: or \: \: 2 - 2 \sqrt{2}$