Question

Is $x=\frac{2}{3}$ a solution of the equation $2x^2-5x-3=0$?

Answer 1

Step 1: State the factor theorem.

Given equation: $2x^2-5x-3=0$.

Let us assume that, $p\left(x\right)=0$, here, $p\left(x\right)=2x^2-5x-3$.

If, $x=a$ is a solution to the equation $p\left(x\right)=0$, then $p\left(a\right)=0$.

Step 2: Apply the factor theorem.

Put $x=\frac{2}{3}$ in the given equation.

$$\begin{gathered} p\left(\frac{2}{3}\right)=2{\left(\frac{2}{3}\right)}^2-5\left(\frac{2}{3}\right)-3 \\ p\left(\frac{2}{3}\right)=2·\frac{4}{9}-\frac{10}{3}-3 \\ p\left(\frac{2}{3}\right)=\frac{8}{9}-\frac{10}{3}-3 \\ p\left(\frac{2}{3}\right)=\frac{8-30-27}{9} \\ p\left(\frac{2}{3}\right)=-\frac{49}{9} \end{gathered}$$

Hence, $p\left(\frac{2}{3}\right)\ne 0$.

Therefore, $x=\frac{2}{3}$ is not a solution to the given equation $2x^2-5x-3=0$.