Question

Determine 'k' such that the quadratic equation $x^2+7(3+2k)-2x(1+3k)=0$ has equal roots.

• A. $2,-10/9$
• B. $3,-10/9$
• C. $2,10/9$
• D. None of these

Answer 1

Answer: (A)

$2,-10/9$

Given equation is: $x^2+7(3+2k)-2x(1+3k)=0$

$\therefore a=1,b=-2(1+3k),c=7(3+2k)$

For equal roots $D=0$, $b^{-}4ac=0$
$\Rightarrow {\left[2(1+3k)\right]}^2-4.7(3+2k)=0$
$\Rightarrow 4(1+6k+{9k}^2)-28(3+2k)=0$
$\Rightarrow 1+6k+{9k}^2-21-14k=0$
$\Rightarrow {9k}^2-8k-20=0$
$\Rightarrow (k-2)(9k+10)=0$
Hence, $k=2,-\frac{10}{9}$