Q. 330 - Intermediate Algebra

Question

In the following exercises, solve.
$\sqrt[3]{x^2-x+18}=\sqrt[3]{2x^2-3x-6}$

Step-by-Step Solution

Answer

The solutions of the given equation are $x=-4$ and $x=6$.

Step 1. Given information.

The given equation is:

$\sqrt[3]{x^2-x+18}=\sqrt[3]{2x^2-3x-6}$

Step 2. Simplify the equation.

Take cube on both sides.

$\begin{array}{rcl}(\sqrt[3]{x^2-x+18}{)}^3& =& (\sqrt[3]{2x^2-3x-6}{)}^3\\ x^2-x+18& =& 2x^2-3x-6\\ 0& =& 2x^2-3x-6-x^2+x-18\\ 0& =& x^2-2x-24\end{array}$

Step 3. Solve the equation.

Splitting the middle term, the above equation can be written as:

$\begin{array}{rcl}{x}^2-6x+4x-24& =& 0\\ x(x-6)+4(x-6)& =& 0\\ (x+4)(x-6)& =& 0\\ x& =& -4,6\end{array}$

Step 4. Conclusion.

The solutions of the given equation are $x=-4$ and $x=6$.