Question

The value of $k$ for which the equation $(k-12)x^2+2(k-12)x+2=0$ has coincident roots is

• A. 13 or -13
• B. 14 or -14
• C. 12 or 14
• D. 16 or 12

Answer 1

The correct option is C 12 or 14

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

For coincident or equal roots, D = 0
i.e $b^2-4ac=0$

Here,
a = (k -12)
b = 2(k - 12)
c = 2

$\Rightarrow \left[2\right(k-12){]}^2-4(k-12)\times 2=0$
$\Rightarrow 2(k-12)\left[2\right(k-12)-4]=0$
$\Rightarrow (k-12)[2k-24-4]=0$
$\Rightarrow (k-12)(2k-28)=0$
$\Rightarrow (k-12)=0\text{or}(2k-28)=0$
$\Rightarrow k=12\text{or}k=14$