Question

The value of $k$ for which the quadratic equation, $k{x}^2+1=kx+3x-11x^2$ has real and equal roots are

• A. -11,-3
• B. 5,7
• C. 5,-7
• D. None of these

Answer 1

The correct option is C 5,-7

Here, we have $k{x}^2+1=kx+3\times -11k^2$

$(k+11)x^2-(k+3)x+1=0$

$a=k+11$

$\Rightarrow b=-(k+3)$ and $c=1$

Since, given equation has real & equal roots, so D = 0

$b^2-4ac=0$

$(-k-3{)}^2-4(k+11\left)\right(1)=0$

$k^2+2k-35=0$

$\Rightarrow (k+7)(k-5)=0$

$k=-7,5$.