Question

If one root of the quadratic equation $k{x}^2-3x-1=0$ is $\frac{1}{2}$, then find the value of $k$.

Answer 1

$k{x}^2-3x-1=0$

One root $=\frac{1}{2}$

$a=k,b=-3,c=-1$

Sum of roots $=\frac{-b}{a}$

Let the other root be $x$

$x+\frac{1}{2}=\frac{3}{k}$

$(2x+1)k=6$

$2xk+k=\mathrm{6.......}\left(1\right)$

Product of roots $=\frac{c}{a}$

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

$xk=-\mathrm{2.......}\left(2\right)$

Solving eq. $\left(1\right)$ and $\left(2\right)$

$2xk+k=62xk=-4-+\_\_\_\_\_\_\_\_\_k=10$