3.6 Solve Radical Equations

Step 1: To isolate the radical, subtract [latex]1[/latex] to both sides.

[latex]\sqrt{9k-2}+1{\color{red}{-}}{\color{red}{1}}=0{\color{red}{-}}{\color{red}{1}}[/latex]

Step 2: Simplify.

[latex]\sqrt{9k-2}=-1[/latex]

Because the square root is equal to a negative number, the equation has no solution.

Step 1: To isolate the radical, subtract [latex]1[/latex] from both sides.

[latex]\sqrt{p-1}+1{\color{red}{-}}{\color{red}{1}}=p{\color{red}{-}}{\color{red}{1}}[/latex]

Step 2: Simplify.

[latex]\sqrt{p-1}=p-1[/latex]

Step 3: Square both sides of the equation.

[latex]\left(\sqrt{p-1}\right)^2=\left(p-1\right)^2[/latex]

Step 4: Simplify, using the Product of Binomial Squares Pattern on the right. Then solve the new equation.

[latex]p-1=p^2-2p+1[/latex]

Step 5: It is a quadratic equation, so get zero on one side.

[latex]0=p^2-3p+2[/latex]

Step 6: Factor the right side.

[latex]0=\left(p-1\right)\left(p-2\right)[/latex]

Step 7: Use the Zero Product Property.

[latex]0=p-1\;\;\;0=p-2[/latex]

Step 8: Solve each equation.

[latex]p=1\;\;\;\;\;\;\;\;\;p=2[/latex]

Step 9: Check the answers.

[latex]\begin{align*} p&={\color{red}{1}}\;\;\;&\;p&={\color{red}{2}}\\ \sqrt{p-1}+1&=p\;\;\;&\;\sqrt{p-1}+1&=p\\ \sqrt{{\color{red}{1}}-1}+1&\overset?={\color{red}{1}}\;\;\;&\;\sqrt{{\color{red}{2}}-1}+1&\overset?={\color{red}{2}}\\ \sqrt{0}+1&\overset?={\color{red}{1}}\;\;\;&\;\sqrt{1}+1&\overset?={\color{red}{2}}\\ 1&=1\checkmark\;\;\;&\;2&=2\checkmark \end{align*}[/latex]

The solutions are [latex]p=1,\;p=2[/latex]

Step 1: To isolate the term with the rational exponent, subtract [latex]3[/latex] from both sides.

[latex]{\left(3x-2\right)}^{\frac{1}{4}}=2[/latex]

Step 2: Raise each side of the equation to the fourth power.

[latex]{\left({\left(3x-2\right)}^{\frac{1}{4}}\right)}^{4}={\left(2\right)}^{4}[/latex]

Step 3: Simplify.

[latex]3x-2=16[/latex]

Step 4: Solve the equation.

[latex]\begin{align*}3x&=18\\x&=6\end{align*}[/latex]

Step 5: Check the answer.

[latex]\begin{align*} x&={\color{red}{6}}\\ (3x-2)^{\frac14}+3&=5\\ (3\cdot{\color{red}{6}}-2)^{\frac14}+3&\overset?=5\\ (16)^{\frac14}+3&\overset?=5\\ 2+3&\overset?=5\\ 5&=5\checkmark \end{align*}[/latex]

The solution is [latex]x=6[/latex].

Step 1: The radical terms are isolated.

[latex]\sqrt[3]{4x-3}=\sqrt[3]{3x+2}[/latex]

Step 2: Since the index is [latex]3[/latex], cube both sides of the equation.

[latex]{\left(\sqrt[3]{4x-3}\right)}^{3}={\left(\sqrt[3]{3x+2}\right)}^{3}[/latex]

Step 3: Simplify, then solve the new equation.

[latex]\begin{align*}4x-3&=3x+2\\x-3&=2\\x&=5\\\text{The solution is}\;x&=5.\end{align*}[/latex]

Step 4: Check the answer.
We leave it to you to show that [latex]5[/latex] checks!

Step 1: Isolate one of the radical terms on one side of the equation.
The radical on the right is isolated.

[latex]\sqrt m+1=\sqrt{m+9}[/latex]

Step 2: Raise both sides of the equation to the power of the index.
We square both sides.

[latex]{(\sqrt m+1)}^2={(\sqrt{m+9})}^2[/latex]

Simplify — be very careful as you multiply!

Step 3: Are there any more radicals?
If yes, repeat Step 1 and Step 2 again. If no, solve the new equation.
There is still a radical in the equation. So we must repeat the previous steps. Isolate the radical term.

[latex]m+2\sqrt m+1=m+9[/latex]

Here, we can easily isolate the radical by dividing both sides by [latex]2[/latex].

[latex]\begin{align*}2\sqrt m&=8\\\sqrt m&=4\end{align*}[/latex]

Square both sides.

[latex]\begin{align*}\left(\sqrt m\right)^2&=\left(4\right)^2\\m&=16\end{align*}[/latex]

Step 4: Check the answer in the original equation.

[latex]\begin{align*}\sqrt {\color{red}{m}}+1&=\sqrt{{\color{red}{m}}+9}\\\sqrt{\color{red}{16}}+1&\overset?=\sqrt{{\color{red}{16}}+9}\\4+1&\overset?=5\\5&=5\checkmark\end{align*}[/latex]

The solution is [latex]m=16[/latex]

Step 1: The radical on the right is isolated. Square both sides.

[latex]\left(\sqrt{q-2}+3\right)^2=\left(\sqrt{4q+1}\right)^2[/latex]

Step 2: Simplify.

[latex]q-2+6\sqrt{q-2}+9=4q+1[/latex]

Step 3: There is still a radical in the equation so we must repeat the previous steps. Isolate the radical.

[latex]6\sqrt{q-2}=3q-6[/latex]

Step 4: Square both sides. It would not help to divide both sides by [latex]6[/latex]. Remember to square both the [latex]6[/latex] and the [latex]\sqrt{q-2}[/latex]

[latex]\begin{align*}\left(6\sqrt{q-2}\right)^2&=\begin{pmatrix}{\color{red}{a}}&{\color{red}{-}}&{\color{red}{b}}\\ 3q&-&6\end{pmatrix}^2\\6^2\left(\sqrt{q-2}\right)^2&=\begin{array}{ccccc}{\color{red}{a^2}}&{\color{red}{-}}&{\color{red}{2ab}}&{\color{red}{+}}&{\color{red}{b^2}}\\\left(3q\right)^2&-&2\cdot3q\cdot6&+&6^2\end{array} \end{align*}[/latex]

Step 5: Simplify, then solve the new equation.

[latex]36\left(q-2\right)=9q^2-36q+36[/latex]

Step 6: Distribute.

[latex]36q-72=9q^2-36q+36[/latex]

Step 7: It is a quadratic equation, so get zero on one side.

[latex]0=9q^2-72q+108[/latex]

Step 8: Factor the right side.

[latex]\begin{align*}0&=9\left(q^2-8q+12\right)\\0&=9\left(q-6\right)\left(q-2\right)\end{align*}[/latex]

Step 9: Use the Zero Product Property.

[latex]\begin{align*}q-6&=0\;\;\;&\;q-2&=0\\ q&=6\;\;\;&\;q&=2 \end{align*}[/latex]

Step 10: The checks are left to you.

The solutions are [latex]q=6[/latex] and [latex]q=2[/latex]

Step 1: Read the problem.

Step 2: Identify what we are looking for.

The time it takes for the sunglasses to reach the river.

Step 3: Name what we are looking.

Let [latex]t=[/latex] time.

Step 4: Translate into an equation by writing the appropriate formula. Substitute in the given information.

[latex]\begin{align*}t&=\frac{\sqrt h}4,\;\text{and}\;h=400\\t&=\frac{\sqrt{\color{red}{400}}}4\end{align*}[/latex]

Step 5: Solve the equation.

[latex]\begin{align*}t&=\frac{20}4\\t&=5\end{align*}[/latex]

Step 6: Check the answer in the problem and make sure it makes sense.

[latex]\begin{align*}5&\overset?=\frac{\sqrt{400}}4\\5&\overset?=\frac{20}4\\5&=5\checkmark\end{align*}[/latex]

Does [latex]5[/latex] seconds seem like a reasonable length of time?

Yes.

Step 7: Answer the question.

It will take [latex]5[/latex] seconds for the sunglasses to reach the river.