Question

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

Answer 1

Since, $x=\frac{-1}{3}$ is a root of the quadratic equation 2x² + 2x + k = 0,
then, it must satisfies the equation.

​$$\begin{gathered} 2{\left(-\frac{1}{3}\right)}^2+2\left(-\frac{1}{3}\right)+k=0 \\ \Rightarrow 2\left(\frac{1}{9}\right)-\frac{2}{3}+k=0 \\ \Rightarrow \frac{2}{9}-\frac{2}{3}+k=0 \\ \Rightarrow \frac{2-6+9k}{9}=0 \\ \Rightarrow -4+9k=0 \\ \Rightarrow 9k=4 \\ \Rightarrow k=\frac{4}{9} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}k\mathrm{is}\frac{4}{9}. \end{gathered}$$