Radical Equations Practice

Notice that this equation contains two radicals, but that each radical is completely isolated on opposite sides of the equation.

We can therefore square both sides to eliminate the radical terms.

$\begin{aligned} \sqrt{y^2+4}&=\sqrt{4y}\\\\\left(\sqrt{y^2+4}\right)^ 2&=\left(\sqrt{4y}\right)^{ {2}}\\\\y^2+4&=4y\end{aligned}$

We obtained a quadratic equation whose solution is $y=2$.

Let's now check for extraneous solutions.

The original equation is ‍$\sqrt{y^2+4}=\sqrt{4y}$. When ${y}= {2}$, we get:

$\begin{aligned}\sqrt{ {y}^2+4}&=\sqrt{4 {y}}\\\\\sqrt{\left( {2}\right)^2+4}&\stackrel{?}{=} \sqrt{4\left( {2}\right)}\\\\\sqrt{8}&=\sqrt{8}\\\\\end{aligned}$

So‍ $y=2$ is a solution.

The answer is: $y=2$

To solve a radical equation, we start by isolating the radical term.

Since the radical term is already isolated, we can square both sides to eliminate the radical:

$\sqrt{12-8w}=6\\\\\left(\sqrt{12-8w}\right)^2=6^{{2}}\\\\12-8w=36$

We obtained a linear equation whose solution is ‍$w=-3$.

Let's now check for extraneous solutions.

The original equation is ‍$\sqrt{12-8w}=6$. When ‍${w}= {-3}$, we get:

$\begin{aligned}\sqrt{12-8 {w}}&=6\\\\\sqrt{12-8( {-3})}&\stackrel{?}{=} 6\\\\\sqrt{12+24}&\stackrel{?}=6\\\\\sqrt{36}&\stackrel{?}=6\\\\6&=6\end{aligned}$

So ‍$w=-3$ is a solution.

The answer is: $w=-3$

To solve a radical equation, we start by isolating the radical term.

Since the radical term is already isolated, we can square both sides to eliminate the radical:

$\begin{aligned} 3w-7=\sqrt{8w-7}\\\\\left(3w-7\right)^2=\left(\sqrt{8w-7}\right)^{{2}}\\\\9w^2-42w+49=8w-7\end{aligned}$

We obtained a quadratic equation whose solution is ‍$w=\dfrac{14}{9}$ and $w=4$.

Let's now check for extraneous solutions.

The original equation is ‍$3w-7=\sqrt{8w-7}$. When ‍${w}= {\dfrac{14}{9}}$, we get:

$\begin{aligned}3 {w}-7&=\sqrt{8 {w}-7}\\\\3\left( {\dfrac{14}{9}}\right)-7&\stackrel{?}{=} \sqrt{8\left( {\dfrac{14}{9}}\right)-7}\\\\-\dfrac{7}{3}&\stackrel{?}=\sqrt{\dfrac{49}{9}}\\\\-\dfrac{7}{3}&\neq \dfrac{7}{3}\end{aligned}$

So ‍$w=\dfrac{14}{9}$ is not a solution.

When $w=4$, we get:

$\begin{aligned}3 {w}-7&=\sqrt{8 {w}-7}\\\\3\left( {4}\right)-7&\stackrel{?}{=} \sqrt{8\left( {4}\right)-7}\\\\5&\stackrel{?}=\sqrt{25}\\\\5&=5\end{aligned}$

So $w=4$ is a solution.

The answer is: $w=4$

Notice that this equation contains two radicals, but that each radical is completely isolated on opposite sides of the equation.

We can therefore square both sides to eliminate the radical terms.

$\begin{aligned} \sqrt{8x-1}&=\sqrt{3-4x}\\\\\left(\sqrt{8x-1}\right)^ 2&=\left(\sqrt{3-4x}\right)^{ {2}}\\\\8x-1&=3-4x\end{aligned}$

We obtained a quadratic equation whose solution is $x=\dfrac{1}{3}$.

Let's now check for extraneous solutions.

The original equation is ‍$\sqrt{8x-1}=\sqrt{3-4x}$. When ${x}= {\dfrac{1}{3}}$, we get:

$\begin{aligned}\sqrt{8 {x}-1}&=\sqrt{3-4 {x}}\\\\\sqrt{8\left( {\dfrac{1}{3}}\right)-1}&\stackrel{?}{=} \sqrt{3-4\left( {\dfrac{1}{3}}\right)}\\\\\sqrt{\dfrac{8}{3}-1}&\stackrel{?}=\sqrt{3-\dfrac{4}{3}}\\\\\sqrt{\dfrac{5}{3}}&=\sqrt{\dfrac{5}{3}}\\\\\end{aligned}$

So‍ $x=\dfrac{1}{3}$ is a solution.

The answer is: $x=\dfrac{1}{3}$