Radical Equations with Two Radicals
Steps for solving radical equations with two radicals:
Example 1: Solve over the real numbers: \(\sqrt{6x+10}=\sqrt{x^2+3x}\).
Steps for solving radical equations with two radicals:
- Make sure the radicals are on opposite sides of the equation.
- Raise each side to the power of the index of the radical.
- Solve the equation.
- Check for extraneous solutions.
Example 1: Solve over the real numbers: \(\sqrt{6x+10}=\sqrt{x^2+3x}\).
\(\begin{align}6x+10&=x^2+3x\ & \ &\text{1) The radicals were already on opposite sides so you can square each side right away.}\\
x^2-3x-10&=0\ & \ &\text{2) Move all terms to one side.}\\
\left(x-5\right)\left(x+2\right)&=0\ & \ &\text{2) Factor.}\\
x&=5,-2\ & \ &\text{4) Solve.}\end{align}\)
x^2-3x-10&=0\ & \ &\text{2) Move all terms to one side.}\\
\left(x-5\right)\left(x+2\right)&=0\ & \ &\text{2) Factor.}\\
x&=5,-2\ & \ &\text{4) Solve.}\end{align}\)
Check:
\(x=5\)
\(\sqrt{6\left(5\right)+10}=\sqrt{5^2+3\left(5\right)}\)
\(\sqrt{40}=\sqrt{40}\) ✓
\(x=-2\)
\(\sqrt{6\left(-2\right)+10}=\sqrt{\left(-2\right)^2+3\left(-2\right)}\)
\(\sqrt{-2}=\sqrt{-2}\) ✘
Since we are solving these equations over the real numbers and the \(\sqrt{-2}\) is not a real number, then \(x=-2\) is an extraneous solution and the only solution is \(x=5\).
Example 2: Solve over the real numbers: \(\sqrt[3]{4x^2+3x+5}-\sqrt[3]{x^2-5x}=0\).
\(x=5\)
\(\sqrt{6\left(5\right)+10}=\sqrt{5^2+3\left(5\right)}\)
\(\sqrt{40}=\sqrt{40}\) ✓
\(x=-2\)
\(\sqrt{6\left(-2\right)+10}=\sqrt{\left(-2\right)^2+3\left(-2\right)}\)
\(\sqrt{-2}=\sqrt{-2}\) ✘
Since we are solving these equations over the real numbers and the \(\sqrt{-2}\) is not a real number, then \(x=-2\) is an extraneous solution and the only solution is \(x=5\).
Example 2: Solve over the real numbers: \(\sqrt[3]{4x^2+3x+5}-\sqrt[3]{x^2-5x}=0\).