Question

Find the value of $K$ for which $4x^2-2(K+1)x+(K+4)=0$ has real and equal roots

Answer 1

Find the value of K

$4x^2-2(K+1)x+(K+4)=0$ root is equal.

$\to$ two root is $\alpha \&\beta$

$\alpha ,\beta =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{+2(K+1)\pm \sqrt{4(K+1{)}^2-4\times 4\times (K+4)}}{2\times 4}$

$=\frac{(2K+2)\pm \sqrt{4(K^2+2K+1)}-16K-64}{8}$

$=\frac{(2K+2)\pm \sqrt{4k^2+8K+4-16K-64}}{8}$

$=\frac{(2K+2)\pm \sqrt{4K^2-8K-60}}{8}$

$\to$ two roots are equal.

$\alpha =\beta$

$\therefore \frac{(2K+2)+\sqrt{4K^2-8K-60}}{8}=\frac{(2K+2)-\sqrt{4K^2-8K-60}}{8}$

$\therefore \sqrt{4K^2-8K-60}=-\sqrt{4K^2-8K-60}$

$\therefore 2\sqrt{4K^2-8K-60=0}$

$\therefore \sqrt{4K^2-8K-60=0}$

$\therefore 4K^2-8K-60=0$

$\therefore 4(K^2-2K-15)=0$

$\therefore K^2-2K-15=0$

$\therefore K^2-5K+3K-15=0$

$\therefore K(K-5)+3(K-5)=0$

$\therefore (K+3)(k-5)=0$

$K=-3$ and $K=5$