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Find the roots of basic equations containing roots

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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

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

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 $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.