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Difficulty Level: At Grade Created by: CK-12
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What if you had a radical equation like 2x+53=0\begin{align*}\sqrt{2x + 5} - 3 = 0\end{align*}? How could you find its real solutions? After completing this Concept, you'll be able to solve radical equations like this one.

### Guidance

When the variable in an equation appears inside a radical sign, the equation is called a radical equation. To solve a radical equation, we need to eliminate the radical and change the equation into a polynomial equation.

A common method for solving radical equations is to isolate the most complicated radical on one side of the equation and raise both sides of the equation to the power that will eliminate the radical sign. If there are any radicals left in the equation after simplifying, we can repeat this procedure until all radical signs are gone. Once the equation is changed into a polynomial equation, we can solve it with the methods we already know.

We must be careful when we use this method, because whenever we raise an equation to a power, we could introduce false solutions that are not in fact solutions to the original problem. These are called extraneous solutions. In order to make sure we get the correct solutions, we must always check all solutions in the original radical equation.

Let’s consider a few simple examples of radical equations where only one radical appears in the equation.

#### Example A

Find the real solutions of the equation 2x1=5\begin{align*}\sqrt{2x-1}=5\end{align*}.

Solution

Since the radical expression is already isolated, we can just square both sides of the equation in order to eliminate the radical sign:

(2x1)2=52\begin{align*}\left(\sqrt{2x-1}\right)^2=5^2\end{align*}

Remember that a2=a so the equation simplifies to:Add one to both sides:Divide both sides by 2:2x12x=25=26x=13\begin{align*}\text{Remember that} \ \sqrt{a^2}=a \ \text{so the equation simplifies to:} && 2x-1& =25\\ \text{Add one to both sides:} && 2x& =26\\ \text{Divide both sides by 2:} &&& \underline{\underline{x=13}}\end{align*}

Finally we need to plug the solution in the original equation to see if it is a valid solution.

2x1=2(13)1=261=25=5\begin{align*}\sqrt{2x-1}=\sqrt{2(13)-1}=\sqrt{26-1}=\sqrt{25}=5\end{align*} The solution checks out.

#### Example B

Find the real solutions of 37x33=0\begin{align*}\sqrt[3]{3-7x}-3=0\end{align*}.

Solution

We isolate the radical on one side of the equation:Raise each side of the equation to the third power:Simplify:Subtract 3 from each side:Divide both sides by -7:37x3(37x3)337x7x=3=33=27=24x=247\begin{align*}\text{We isolate the radical on one side of the equation:} && \sqrt[3]{3-7x}& =3\\ \text{Raise each side of the equation to the third power:} && \left(\sqrt[3]{3-7x}\right)^3& =3^3\\ \text{Simplify:} && 3-7x& =27\\ \text{Subtract 3 from each side:} && -7x& =24\\ \text{Divide both sides by -7:} &&& \underline{\underline{x=-\frac{24}{7}}}\end{align*}

Check: \begin{align*}\sqrt[3]{3-7x}-3=\sqrt[3]{3-7 \left(-\frac{24}{7}\right)}-3=\sqrt[3]{3+24}-3=\sqrt[3]{27}-3=3-3=0\end{align*}. The solution checks out.

#### Example C

Find the real solutions of \begin{align*}\sqrt{10-x^2}-x=2\end{align*}.

Solution

\begin{align*}\text{We isolate the radical on one side of the equation:} && \sqrt{10-x^2}& =2+x\\ \text{Square each side of the equation:} && \left(\sqrt{10-x^2}\right)^2& =(2+x)^2\\ \text{Simplify:} && 10-x^2& =4+4x+x^2\\ \text{Move all terms to one side of the equation:} && 0& =2x^2+4x-6\\ \text{Solve using the quadratic formula:} && x& =\frac{-4 \pm \sqrt{4^2-4(2)(-6)}}{4}\\ \text{Simplify:} && x& =\frac{-4 \pm \sqrt{64}}{4}\\ \text{Re-write} \ \sqrt{24} \ \text{in simplest form:} && x& =\frac{-4 \pm 8}{4}\\ \text{Reduce all terms by a factor of 2:} && x& =1 \ \text{or} \ x=-3\end{align*}

Check: \begin{align*}\sqrt{10-1^2}-1=\sqrt{9}-1=3-1=2\end{align*} This solution checks out.

\begin{align*}\sqrt{10-(-3)^2}-(-3)=\sqrt{1}+3=1+3=4\end{align*} This solution does not check out.

The equation has only one solution, \begin{align*}\underline{\underline{x=1}}\end{align*}; the solution \begin{align*}x=-3\end{align*} is extraneous.

#### Example D

A sphere has a volume of \begin{align*}456 \ cm^3\end{align*}. If the radius of the sphere is increased by 2 cm, what is the new volume of the sphere?

Solution

Make a sketch:

Define variables: Let \begin{align*}R =\end{align*} the radius of the sphere.

Find an equation: The volume of a sphere is given by the formula \begin{align*}V=\frac{4}{3}\pi R^3\end{align*}.

Solve the equation:

\begin{align*}\text{Plug in the value of the volume:} && 456& =\frac{4}{3} \pi R^3\\ \text{Multiply by 3:} && 1368& =4 \pi R^3\\ \text{Divide by} \ 4 \pi: && 108.92& =R^3\\ \text{Take the cube root of each side:} && R& =\sqrt[3]{108.92} \Rightarrow R=4.776 \ cm\\ \text{The new radius is 2 centimeters more:} && R& =6.776 \ cm\\ \text{The new volume is:} && V & =\frac{4}{3} \pi (6.776)^3=\underline{\underline{1302.5}} \ cm^3\end{align*}

Check: Let’s plug in the values of the radius into the volume formula:

\begin{align*}V=\frac{4}{3} \pi R^3=\frac{4}{3} \pi (4.776)^3=456 \ cm^3\end{align*}. The solution checks out.

#### Example E

The kinetic energy of an object of mass \begin{align*}m\end{align*} and velocity \begin{align*}v\end{align*} is given by the formula: \begin{align*}KE=\frac{1}{2} mv^2\end{align*}. A baseball has a mass of 145 kg and its kinetic energy is measured to be 654 Joules \begin{align*}(kg \cdot m^2/s^2)\end{align*} when it hits the catcher’s glove. What is the velocity of the ball when it hits the catcher’s glove?

Solution

\begin{align*}\text{Start with the formula:} && KE& =\frac{1}{2} mv^2\\ \text{Plug in the values for the mass and the kinetic energy:} && 654 \frac{kg \cdot m^2}{s^2}& =\frac{1}{2}(145\ kg)v^2\\ \text{Multiply both sides by 2:} && 1308 \frac{kg \cdot m^2}{s^2}& =145 \ kg \cdot v^2\\ \text{Divide both sides by 145} \ kg: && 9.02 \frac{m^2}{s^2}& =v^2\\ \text{Take the square root of both sides:} && v& =\sqrt{9.02} \sqrt{\frac{m^2}{s^2}}=3.003 \ m/s\end{align*}

Check: Plug the values for the mass and the velocity into the energy formula:

\begin{align*}KE=\frac{1}{2}mv^2=\frac{1}{2}(145 \ kg)(3.003 \ m/s)^2=654 \ kg \cdot m^2/s^2\end{align*}

Watch this video for help with the Examples above.

### Vocabulary

• For a quadratic equation in standard form, \begin{align*}ax^2 + bx + c = 0\end{align*}, the quadratic formula looks like this:

\begin{align*}x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\end{align*}

### Guided Practice

Find the real solutions of \begin{align*}\sqrt{3x-9}-1=5\end{align*}.

Solution

\begin{align*}\text{We isolate the radical on one side of the equation:} && \sqrt{3x-9}&=6\\ \text{Square each side of the equation:} && \sqrt{3x-9}^2&=6^2\\ \text{Simplify:} && 3x-9&=36\\ \text{Add 9 from each side:} && 3x&=45\\ \text{Divide both sides by 3:} &&& \underline{\underline{x=\frac{45}{3}=15}}\end{align*}

Check: \begin{align*}\sqrt{3x-9}-1=\sqrt{3(15)-9}-1=\sqrt{45-9}-1=\sqrt{36}-1=6-1=5.\end{align*} The solution checks out.

### Practice

Find the solution to each of the following radical equations.

1. \begin{align*}\sqrt{x+2}-2=0\end{align*}
2. \begin{align*}\sqrt{3x-1}=5\end{align*}
3. \begin{align*}2 \sqrt{4-3x}+3=0\end{align*}
4. \begin{align*}\sqrt[3]{x-3}=1\end{align*}
5. \begin{align*}\sqrt[4]{x^2-9}=2\end{align*}
6. \begin{align*}\sqrt[3]{-2-5x}+3=0\end{align*}
7. \begin{align*}\sqrt{x^2-3}=x-1\end{align*}
8. \begin{align*}\sqrt{x}=x-6\end{align*}
9. \begin{align*}\sqrt{x^2-5x}-6=0\end{align*}
10. \begin{align*}\sqrt{(x+1)(x-3)}=x\end{align*}
11. \begin{align*}\sqrt{x+6}=x+4\end{align*}
12. \begin{align*}\sqrt{3x+4}=-6\end{align*}
13. The area of a triangle is \begin{align*}24 \ in^2\end{align*} and the height of the triangle is twice as long as the base. What are the base and the height of the triangle?
14. The length of a rectangle is 7 meters less than twice its width, and its area is \begin{align*}660 \ m^2\end{align*}. What are the length and width of the rectangle?
15. The area of a circular disk is \begin{align*}124 \ in^2\end{align*}. What is the circumference of the disk? \begin{align*}(\text{Area} = \pi R^2, \text{Circumference} =2 \pi R)\end{align*}.
16. The volume of a cylinder is \begin{align*}245 \ cm^3\end{align*} and the height of the cylinder is one third of the diameter of the base of the cylinder. The diameter of the cylinder is kept the same but the height of the cylinder is increased by 2 centimeters. What is the volume of the new cylinder? \begin{align*}(\text{Volume} =\pi R^2 \cdot h)\end{align*}
17. The height of a golf ball as it travels through the air is given by the equation \begin{align*}h=-16t^2+256\end{align*}. Find the time when the ball is at a height of 120 feet.

### Vocabulary Language: English

Extraneous Solution

Extraneous Solution

An extraneous solution is a solution of a simplified version of an original equation that, when checked in the original equation, is not actually a solution.

A quadratic equation is an equation that can be written in the form $=ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real constants and $a\ne 0$.

The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

A radical expression is an expression with numbers, operations and radicals in it.

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Date Created:
Oct 01, 2012