This topic puts all the radical rules together and helps solve equation that have radicals. Depending on the type of radical in the equation, there may be one or multiple solutions from the algebra. Checking each one is key to determining the correct solution(s). REMEMBER: check proposed solutions into the original equation.
Example Solve the equation: \(\sqrt{5x+1}=4\)
In the previous example, the approach was to square both sides of the equation. This eliminates the square root from the equation, creating an equation we already know how to solve. In general, whatever index the radical has, that is the power that both sides of the equation needs to be raised to:
$$\sqrt[\color{red}{n}]{\text{expression}}=\text{expression}\implies\left(\sqrt[n]{\text{expression}}\right)^{\color{red}{n}}=\left(\text{expression}\right)^{\color{red}{n}}$$Example Solve the equation: \(\sqrt[5]{2x+7}=\sqrt[5]{3x-2}\)
| \(\sqrt[5]{2x+7}=\sqrt[5]{3x-2}\) | \(\implies\left(\sqrt[5]{2x+7}\right)^5=\left(\sqrt[5]{3x-2})\right)^5\) |
| \(\implies2x+7=3x-2\implies x=9\) | |
| Check: | \(\sqrt[5]{2(9)+7}=\sqrt[5]{3(9)-2}\implies\sqrt[5]{25}=\sqrt[5]{25}\;\checkmark\) |
In the following examples, some algebra needs to be applied to get the radical by itself first. Then apply to technique used in the previous examples. Since these are equations, watch out for NO SOLUTION and INFINITE SOLUTIONS as well. It might take until the checking proposed solutions part to see these possibilities.
Example Solve \(\sqrt{3x-5}=x-1\).
Example Solve \(\sqrt{4-x}-x=2\)