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. $\{-5,-7\}$

Answer 1

The correct option is C

$\{5,-7\}$

Explanation for the correct option:

In the question, an equation $k{x}^2+1=kx+3x-11x^2$ is given.

Rewrite the given equation as follows:

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

Since, it is given that the given equation has real and equal roots.

We know that the discriminant of the quadratic equation which has real and equal roots is always zero.

The formula of discriminant is $D=b^2-4ac$.

Where, $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$, and $c$ is the constant term.

$$\begin{gathered} {(-(k+3))}^2-4\times (k+11\times 1)=0 \\ \Rightarrow k^2+9+6k-4k-44=0 \\ \Rightarrow k^2+2k-35=0 \\ \Rightarrow (k+7)(k-5)=0 \\ \Rightarrow k=\{-7,5\} \end{gathered}$$

Therefore, the value of $k$ are $\{-7,5\}$.

Hence, option C is the correct answer.