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## Find the roots of basic equations containing roots

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Solving Simple Radical Equations

The legs of a right triangle measure 3 and $2\sqrt {x}$ . The hypotenuse measures 5. What is the length of the leg with the unknown value?

### Guidance

Solving radical equations are very similar to solving other types of equations. The objective is to get $x$ by itself. However, now there are radicals within the equations. Recall that the opposite of the square root of something is to square it.

#### Example A

Is $x = 5$ the solution to $\sqrt{2x+15}=8$ ?

Solution: Plug in 5 for $x$ to see if the equation holds true. If it does, then 5 is the solution.

$\sqrt{2 \left(5\right)+15}&=8 \\\sqrt{10+15}&=9 \\\sqrt{25} &\neq 8$

We know that $\sqrt{25}=5$ , so $x = 5$ is not the solution.

#### Example B

Solve $\sqrt{2x-5}+7=16$ .

Solution: To solve for $x$ , we need to isolate the radical. Subtract 7 from both sides.

$\sqrt{2x-5}+7&=16 \\\sqrt{2x-5}&=9$

Now, we can square both sides to eliminate the radical. Only square both sides when the radical is alone on one side of the equals sign.

$\sqrt{2x-5}^2&=9^2 \\2x-5&=81 \\2x&=86 \\x&=43$

Check: $\sqrt{2 \left(43\right)-5}+7=\sqrt{86-5}+7=\sqrt{81}+7=9+7=16$

ALWAYS check your answers when solving radical equations. Sometimes, you will solve an equation, get a solution, and then plug it back in and it will not work. These types of solutions are called extraneous solutions and are not actually considered solutions to the equation.

#### Example C

Solve $3\sqrt[3]{x-8}-2=-14$ .

Solution: Again, isolate the radical first. Add 2 to both sides and divide by 3.

$3\sqrt[3]{x-8}-2&=-14\\3\sqrt[3]{x-8}&=-12\\\sqrt[3]{x-8}&=-4$

Now, cube both sides to eliminate the radical.

$\sqrt[3]{x-8}^3&=(-4)^3\\x-8&=-64\\x&=-56$

Check: $3\sqrt[3]{-56-8}-2=3 \sqrt[3]{-64}-2=3 \cdot -4-2=-12-2=-14$

Intro Problem Revisit Use the Pythagorean Theorem and solve for x then substitute that value in to solve for the leg with the unknown.

$3^2 +(2\sqrt {x})^2) = 5^2\\9 + 4x = 25\\4x = 16\\x = 4$

Now substitute this value into the leg with the unknown.

$2 \sqrt{4} = 2 \cdot 2 = 4$

Therefore the leg with the unknown has a length of 4.

### Guided Practice

Solve the equations and check your answers.

1. $\sqrt{x+5}=6$

2. $5\sqrt{2x-1}+1=26$

3. $\sqrt[4]{3x+11}-2=3$

1. The radical is already isolated here. Square both sides and solve for $x$ .

$\sqrt{x+5}^2&=6^2 \\x+5&=36 \\x&=31$

Check: $\sqrt{31+5}=\sqrt{36}=6$

2. Isolate the radical by subtracting 1 and then dividing by 5.

$5\sqrt{2x-1}+1&=26 \\5\sqrt{2x-1}&=25 \\\sqrt{2x-1}&=5$

Square both sides and continue to solve for $x$ .

$\sqrt{2x-1}^2&=5^2 \\2x-1&=25 \\2x&=26 \\x&=13$

Check: $5\sqrt{2 \left(13\right)-1}+1=5\sqrt{26-1}=5\sqrt{25}+1=5 \cdot 5+1=25+1=26$

3. In this problem, we have a fourth root. That means, once we isolate the radical, we must raise both sides to the fourth power to eliminate it.

$\sqrt[4]{3x+11}-2&=3\\\sqrt[4]{3x-11}^4&=5^4\\3x-11&=625\\3x&=636\\x&=212$

Check: $\sqrt[4]{3 \left(212\right)+11}-2=\sqrt[4]{636-11}-2=\sqrt[4]{625}-2=5-2=3$

### Vocabulary

Extraneous Solution
A solved-for value of $x$ , that when checked, is not actually a solution.

### Practice

Determine if the given values of x are solutions to the radical equations below.

1. $\sqrt{x-3}=7; x = 32$
2. $\sqrt[3]{6+x}=3; x = 21$
3. $\sqrt[4]{2x+3}-11=-9; x = 6$

Solve the equations and check your answers.

1. $\sqrt{x+5}=6$
2. $2- \sqrt{x+1}=0$
3. $4 \sqrt{5-x}=12$
4. $\sqrt{x+9}+7=11$
5. $\frac{1}{2}\sqrt[3]{x-2}=1$
6. $\sqrt[3]{x+3}+5=9$
7. $5\sqrt{15-x}+2=17$
8. $-5=\sqrt[5]{x-5}-7$
9. $\sqrt[4]{x-6}+10=13$
10. $\frac{8}{5}\sqrt[3]{x+5}=8$
11. $3 \sqrt{x+7}-2=25$
12. $\sqrt[4]{235+x}+9=14$

### 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.