Question

A value of $k$ for which the quadratic equation $x^2-2x\left(1+3k\right)+7\left(2k+3\right)=0$ has equal roots, is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$2$

Explanation for correct answer:

Finding the value of k.

Step-1: Apply condition for two roots are common.

Given, quadratic equation $x^2-2x\left(1+3k\right)+7\left(2k+3\right)=0$ has equal roots.

It means discriminant$D$ will be zero.$b^2-4ac=0$

$\Rightarrow$ ${\left(2\left(1+3k\right)\right)}^2-4\times 1\times \left(7\left(2k+3\right)\right)=0$

$\Rightarrow$ $4\left[\left({\left(1\right)}^2+9k^2+6k\right)-\left(14k+21\right)\right]=0$

$\Rightarrow$ $\left[9k^2-8k-20\right]=0.......\left(i\right)$

Step-2: Finding the roots of quadratic equation of equation $\left(i\right)$.

$\Rightarrow$ $k=\frac{-b\pm \sqrt{D}}{2a}$

$\Rightarrow$ $k=\frac{-\left(-8\right)\pm \sqrt{64-\left(4\times 9\times \left(-20\right)\right)}}{2\times 9}$

$\Rightarrow$ $k=\frac{8\pm 28}{18}$

$\Rightarrow$ $k=2,-\frac{10}{9}$

Hence, option $\left(B\right)$ is correct.