Question

If $\frac{1}{2}$ is a root of the equation $x^2+kx-\frac{5}{4}=0$, then the value of k is ________.

Answer 1

Let f(x) = $x^2+kx-\frac{5}{4}$

It is given that, $\frac{1}{2}$ is a root of the equation $x^2+kx-\frac{5}{4}=0$.

Therefore, $f\left(x\right)=0\mathrm{when}x=\frac{1}{2}$.

On putting x = $\frac{1}{2}$ in f(x) = 0, we get

$$\begin{gathered} {\left(\frac{1}{2}\right)}^2+k\left(\frac{1}{2}\right)-\frac{5}{4}=0 \\ \Rightarrow \frac{1}{4}+k\left(\frac{1}{2}\right)-\frac{5}{4}=0 \\ \Rightarrow \frac{1+2k-5}{4}=0 \\ \Rightarrow 2k-4=0 \\ \Rightarrow 2k=4 \\ \Rightarrow k=2 \end{gathered}$$

Hence, if $\frac{1}{2}$ is a root of the equation $x^2+kx-\frac{5}{4}=0$, then the value of k is 2.