Question

Find the value of $k$ for which given equation has real and equal roots.

$(k-12)x^2+2(k-12)x+2=0$

• A. $k=12$
• B. $k=13$
• C. $k=14$
• D. $k=15$

Answer 1

Answer: (C)

$k=14$

$(k-12)x^2-2(k-12)x+2=0$
Since, the roots are real and equal. The discriminant is zero.
$D=b^2-4ac=0$
$(-2(k-12){)}^2-4(k-12\left)\right(2)=0$
$4(k-12{)}^2-8k+96=0$
$4k^2-96k+576-8k+96=0$
$4k^2-104k+672=0$
$k^2-26k+168=0$
$k^2-14k-12k+168=0$
$(k-14)(k-12)=0$
$k=12,14$
Neglect $k=12$ because the equation will not be quadratic if we substitute this value of $k$.
Hence, $k=14$