Question

find the value of k for which the equation 2x2-(k-2)x+1=0 has real and equal roots

Answer 1

For a quadratic equation, roots are real and equal if it's discriminant is 0
${(k - 2)}^{2} - 4 \times 2 \times 1 = 0 \\ {(k - 2)}^{2} = 8 \\ k - 2 = + \: \: or \: \: - 2 \sqrt{2} \\ k = 2 + 2 \sqrt{2} \: \: or \: \: 2 - 2 \sqrt{2}$

Answer 2

given that 2x^2-(k-2)x+1=0
if roots are equal then b^2-4ac=0
so,(k-2)^2-4×2×1=0
=>(k-2)^2=8
=>(k-2)=+or-2√2
k=2+or-2√2