Q. 602 - Intermediate Algebra

Question

In the following exercises, solve.

$\sqrt[3]{2x^2-6x-23}=\sqrt[3]{x^2-3x+5}$

Step-by-Step Solution

Answer

The solution of the equation is $x=7,-4$

Step 1. Given Information

We have $\sqrt[3]{2x^2-6x-23}=\sqrt[3]{x^2-3x+5}$

Step 2. Cubing both sides

$$\begin{gathered} {\left(\sqrt[3]{2x^2-6x-23}\right)}^3={\left(\sqrt[3]{x^2-3x+5}\right)}^3 \\ 2x^2-6x-23=x^2-3x+5 \end{gathered}$$

$x^2-3x-28=0$

Step 3. Factorize the expression

$$\begin{gathered} x^2-3x-28=0 \\ \left(x-7\right)\left(x+4\right)=0 \\ x=7,-4 \end{gathered}$$