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

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

Here, $a=1,b=-2(1+3k)andc=7(3+2k)$

$\therefore D=b^2-4ac$

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

$\Rightarrow D=4(9k^2+6k+1-21-14k)$

$\Rightarrow D=4(9k^2-8k-20)$

The given equation will have equal roots, if

$D=0$

$\Rightarrow 4(9k^2-8k-20)=0$

$\Rightarrow 9k^2-8k-20=0$

$\Rightarrow 9k^2-18k+10k-20=0$

$\Rightarrow (k-2)(9k+10)=0$

$\Rightarrow k-2=0or,9k+10=0\Rightarrow k=2or,k=-\frac{10}{9}$