Question

Find the values of $k$ for which the given equation has real and equal roots
$x^2-2x(1+3k)+7(3+2k)=0$

Answer 1

On comparing given equation with $a{x}^2+bx+c=0$ we get,

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

For the roots to be real and equation $D$ must be equal to zero,

$\begin{array}{cc}D& =b^2-4ac\\ 0& =4(1+3k{)}^2-4\times \left[7\right(3+2k\left)\right]\\ & =9k^2+6k+1-21-14k\\ 9k^2-8k-20& =0\end{array}$

The given equation will have equal roots, if

$\begin{array}{cc}9k^2-8k-20& =0\\ (k-2)(9k+10)& =0\\ k-2=0\text{and}9k+10& =0\\ k=2\text{and}k& =-\frac{10}{9}\end{array}$

Hence, for $k=2$ and $k=-\frac{10}{9}$ given equation will have real and equal roots.