Question

Find the value of $K$ if $\frac{1}{2}$ in a root of the equation $x^2+kx-\frac{s}{4}=0$ .

Answer 1

$x^2+kx-\frac{s}{4}=0$ $x=\frac{1}{2}$

$\frac{1}{4}+\frac{k}{2}-\frac{s}{4}=0\Rightarrow 1+2k-s=0$

$s=1+2k$

$x^2+kx-\frac{1+2k}{4}=0$

$4x^2+4kx-(1+2k)=0\Rightarrow$

$x=\frac{-4k\pm \sqrt{16k^2+16(1+2k)}}{2x4}$

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

$\Rightarrow x=\frac{-4k\pm (4k+4{)}^2}{8}$

$\Rightarrow x=\frac{-k+k+1}{2}$

$=\frac{1}{2}$

& $x=\frac{-k-k-1}{2}$

$x=\frac{-2k-1}{2}$

$\frac{-2k-1}{x}=\frac{1}{2}$

$k=-1$