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Section 6.5 Radical Equations

Subsection 1. Solve radical equations

To solve a simple radical equation, we raise both sides to the index of the radical.

Subsubsection Examples

Example 6.59.

Solve \(~\sqrt{x-3} = 4\)

Solution

We square both sides of the equation to produce an equation without radicals.

\begin{align*} \left(\sqrt{x-3}\right)^2 \amp = 4^2\\ x-3 \amp = 16\\ x \amp = 19 \end{align*}

You can check that \(x=19\) satisfies the original equation.

Example 6.60.

Solve \(~-2\sqrt[3]{x-4} = -6\)

Solution

We first isolate the cube root.

\begin{align*} -2\sqrt[3]{x-4} \amp = -6 \amp \amp \blert{\text{Divide both sides by}~-2.}\\ \sqrt[3]{x-4} \amp = 3 \end{align*}

Next, we undo the cube root by cubing both sides of the equation.

\begin{align*} \left(\sqrt[3]{x-4}\right)^3 \amp = 3^3\\ x-4 \amp = 27 \end{align*}

Finally, we add 4 to both sides to find the solution, \(x=31\text{.}\) We do not have to check for extraneous solutions when we cube both sides of an equation, but it is a good idea to check the solution for accuracy anyway.

Check: We substitute \(\alert{31}\) for \(x\) into the left side of the equation.

\begin{equation*} 2\sqrt[3]{\alert{31}-4} = -2\sqrt[3]{27} = -2(3) = -6 \end{equation*}

The solutions checks.

Subsubsection Exercises

Solve \(~\sqrt{x-6} = 2\)

Answer
\(x=10\)

Solve \(~3\sqrt[3]{4x-1} = -15\)

Answer
\(x=-31\)

Subsection 2. Square binomials containing radicals

We may encounter binomials when squaring both sides of an equation.

Subsubsection Examples

Example 6.63.

Expand \(~(\sqrt{x}-3)^2\)

Solution

\((\sqrt{x}-3)^2 = (\sqrt{x}-3)(\sqrt{x}-3)\text{,}\) so we apply "FOIL" to get

\begin{align*} (\sqrt{x}-3)(\sqrt{x}-3) \amp = \sqrt{x}\sqrt{x}-3\sqrt{x}-3\sqrt{x}-3(-3) \amp \amp \blert{\text{Simplify.}}\\ \amp = x-6\sqrt{x}+9 \end{align*}
Example 6.64.

Expand \(~(8+\sqrt{t-2})^2\)

Solution

We multiply \(~(8+\sqrt{t-2})(8+\sqrt{t-2})~\) to get

\begin{align*} 8\cdot 8+8 \amp \sqrt{t-2}+8\sqrt{t-2}+\left(\sqrt{t-2}\right)\left(\sqrt{t-2}\right)\\ \amp = 64+\sqrt{t-2}+(t+2)\\ \amp = 66+16\sqrt{t-2}+t \end{align*}

Subsubsection Exercises

Expand \(~\left(6-\sqrt{3a+1}\right)^2\)

Answer
\(37-12\sqrt{3a+1}+3a\)

Expand \(~\left(2\sqrt{z+4}-5\right)^2\)

Answer
\(4z+41-20\sqrt{z+4}\)

Subsection 3. Use absolute value

Subsubsection Examples

Example 6.67.

Explain why \(\sqrt{x^2}=x\) is not true for all values of \(x\text{.}\)

Solution

Recall that the symbol \(\sqrt{a}\) means the non-negative square root of \(a\text{.}\) If \(x\) is a negative number, for example \(x=-6\text{,}\) then \(x^2=(-6)^2=36\text{,}\) and not \(\sqrt{x^2}=\sqrt{36}=6\text{,}\) not \(-6\text{.}\) So if \(x\) is a negative number, \(\sqrt{x^2}\not= x\) In fact, \(\sqrt{x^2}= \abs{x}\text{.}\)

Example 6.68.

For what values of \(x\) is \(~\sqrt{(x-5)^2} = x-5~\text{?}\)

Solution

\(~\sqrt{(x-5)^2} = x-5~\) when \(x-5\) is positive or zero, that is for \(x \ge 5\text{.}\) If \(x \lt 5\text{,}\) then \(x-5\) is negative. But the \(\sqrt{}\) symbol returns only the positive root, so we use absolute value bars to indicate that the root is positive:

\begin{equation*} \sqrt{(x-5)^2} = \abs{x-5} \end{equation*}

Subsubsection Exercises

For what values of \(x\) is \(~\sqrt{(2x+8)^2} = 2x+8~\text{?}\)

Answer
\(x \ge -4\)

For what values of \(x\) is \(~\sqrt{(x-9)^2} = 9-x~\text{?}\)

Answer
\(x \le 9\)