## How to Solve Radical Equations

Radical equations are those that contain radical expressions. Applying exponentiation rules and some basic algebraic laws are necessary for solving radical equations.

### Solving Radical Equations

A simple way to solve radical equations is to first isolate the radical term and then raise both sides to a power to get rid of the radical. This is the same way you solved other equations that didn't have radicals: you rearrange the expression to find the variable you want to know, and then you solve the equation that's left.

To solve radical equations, you will need to keep in mind two main ideas.

• The first one is if $$a \ = \ b$$, $$a^2 \ = \ b^2$$. (Because of this property, you can square both sides of an equation and be sure that they are still equal.)
• The second thing is that if you square the square root of any positive number $$x$$, you get $$x$$: $$(\sqrt{x})^2 \ = \ x$$. (With this property, you can get rid of the radicals in your equations.)

Let's start with a radical equation that's easy to solve: $$\sqrt{x} \ + \ 4 \ = \ 6$$

Step1: Minus $$4$$ from both sides of the equation to isolate the variable term on the left side.
$$\sqrt{x} \ + \ 4 \ = \ 6 \ ⇒ \ \sqrt{x} \ = \ 2$$

Step2: Square both sides to get rid of the radical. Then simplify.
$$\sqrt{x} \ = \ 2 \ ⇒ \ (\sqrt{x})^2 \ = \ 4 \ ⇒ \ x \ = \ 4$$

In the example above, the only thing that was under the radical was the variable $$x$$. Sometimes you'll need to figure out how to solve an equation that has more than one term under a radical. Follow the same steps to solve these, but pay attention to a key point: square both sides of an equation, not the individual terms. See how the next example is solved.

### Example

Solve for $$x$$: $$\sqrt{x \ - \ 2} \ + \ 6 \ = \ 10$$

Solution:

Step1: Isolate the variable terms on one side, to do that minus $$6$$ from both sides:
$$\sqrt{x \ - \ 2} \ + \ 6 \ = \ 10 \ ⇒ \ \sqrt{x \ - \ 2} \ = \ 4$$

Step2: To remove the radical, square both sides:
$$(\sqrt{x \ - \ 2})^2 \ = \ 4^2 \ ⇒ \ x \ - \ 2 \ = \ 16$$

Step3: Now, make the equation as simple as possible and find $$x$$:
$$x \ - \ 2 \ = \ 16 \ ⇒ \ x \ = \ 18$$

Step4: Check your answer. If you replace $$x$$ in the original equation with $$18$$, you get a true statement, so the solution is right:
$$\sqrt{x \ - \ 2} \ + \ 6 \ = \ 10 \ ⇒ \ \sqrt{18 \ - \ 2} \ + \ 6 \ = \ 10 \ ⇒ \ \sqrt{16} \ + \ 6 \ = \ 10 \ ⇒ \ 4 \ + \ 6 \ = \ 10$$

### Exercises for Solving Radical Equations

1) Solve for $$x$$: $$\sqrt{x \ - \ 2} \ = \ \sqrt{2x \ + \ 3}$$

2) Solve for $$x$$: $$\sqrt{7x \ + \ 3} \ = \ \sqrt{3x \ + \ 6}$$

3) Solve for $$q$$: $$\sqrt{-3q \ + \ 4} \ = \ \sqrt{5q \ - \ 12}$$

4) Solve for $$m$$: $$\sqrt{m^2 \ - \ 8} \ = \ \sqrt{-3m^2 \ + \ 16}$$

5) Solve for $$z$$: $$\sqrt{-2z^2 \ + \ 12} \ = \ \sqrt{6z^2 \ - \ 20}$$

6) Solve for $$t$$: $$\sqrt{-7t \ + \ 25} \ = \ \sqrt{9t \ - \ 39}$$

7) Solve for $$n$$: $$\sqrt{5n \ + \ 13} \ = \ \sqrt{-6n \ - \ 20}$$

8) Solve for $$z$$: $$\sqrt{-2z^2 \ + \ 12} \ = \ \sqrt{6z^2 \ - \ 20}$$

9) Solve for $$x$$: $$\sqrt{-9x \ + \ 27} \ = \ \sqrt{15x \ - \ 13}$$

10) Solve for $$n$$: $$\sqrt{-7n^2 \ + \ 26} \ = \ \sqrt{9n^2 \ - \ 18}$$

1) Solve for $$x$$: $$\sqrt{x \ - \ 2} \ = \ \sqrt{2x \ + \ 3}$$

$$\color{red}{\sqrt{x \ - \ 2} \ = \ \sqrt{2x \ + \ 3} \ ⇒ \ x \ - \ 2 \ = \ 2x \ + \ 3 \ ⇒ \ -5 \ = \ x}$$

2) Solve for $$x$$: $$\sqrt{7x \ + \ 3} \ = \ \sqrt{3x \ + \ 6}$$

$$\color{red}{\sqrt{7x \ + \ 3} \ = \ \sqrt{3x \ + \ 6} \ ⇒ \ 7x \ + \ 3 \ = \ 3x \ + \ 6 \ ⇒ \ 4x \ = \ 3 \ ⇒ \ x \ = \ \frac{3}{4}}$$

3) Solve for $$q$$: $$\sqrt{-3q \ + \ 4} \ = \ \sqrt{5q \ - \ 12}$$

$$\color{red}{\sqrt{-3q \ + \ 4} \ = \ \sqrt{5q \ - \ 12} \ ⇒ \ -3q \ + \ 4 \ = \ 5q \ - \ 12 \ ⇒ \ 16 \ = \ 8q \ ⇒ \ q \ = \ 2}$$

4) Solve for $$m$$: $$\sqrt{m^2 \ - \ 8} \ = \ \sqrt{-3m^2 \ + \ 16}$$

$$\color{red}{\sqrt{m^2 \ - \ 8} \ = \ \sqrt{-3m^2 \ + \ 16} \ ⇒ \ m^2 \ - \ 8 \ = \ -3m^2 \ + \ 16 \ ⇒ \ 4m^2 \ = \ 24 \ ⇒ \ m^2 \ = \ 6}$$ $$\color{red}{⇒ \ m \ = \ ± \sqrt{6}}$$

5) Solve for $$z$$: $$\sqrt{-2z^2 \ + \ 12} \ = \ \sqrt{6z^2 \ - \ 20}$$

$$\color{red}{\sqrt{-2z^2 \ + \ 12} \ = \ \sqrt{6z^2 \ - \ 20} \ ⇒ \ -2z^2 \ + \ 12 \ = \ 6z^2 \ - \ 20 \ ⇒ \ 32 \ = \ 8z^2 \ ⇒}$$ $$\color{red}{ \ z^2 \ = \ 4 \ ⇒ \ z \ = \ ±2}$$

6) Solve for $$t$$: $$\sqrt{-7t \ + \ 25} \ = \ \sqrt{9t \ - \ 39}$$

$$\color{red}{\sqrt{-7t \ + \ 25} \ = \ \sqrt{9t \ - \ 39} \ ⇒ \ -7t \ + \ 25 \ = \ 9t \ - \ 39 \ ⇒ \ 64 \ = \ 16t \ ⇒ \ t \ = \ 4}$$

7) Solve for $$n$$: $$\sqrt{5n \ + \ 13} \ = \ \sqrt{-6n \ - \ 20}$$

$$\color{red}{\sqrt{5n \ + \ 13} \ = \ \sqrt{-6n \ - \ 20} \ ⇒ \ 5n \ + \ 13 \ = \ -6n \ - \ 20 \ ⇒ \ 11n \ = \ -33 \ ⇒ \ n \ = \ -3}$$

8) Solve for $$z$$: $$\sqrt{-2z^2 \ + \ 12} \ = \ \sqrt{6z^2 \ - \ 20}$$

$$\color{red}{\sqrt{z^2 \ + \ 11z \ - \ 15} \ = \ \sqrt{2z^2 \ - \ 7z \ + \ 12} \ ⇒ \ z^2 \ + \ 11z \ - \ 15 \ = \ 2z^2 \ - \ 7z \ + \ 12}$$ $$\color{red}{⇒ \ -3z^2 \ + \ 18z \ - \ 27 \ = \ 0 \ ⇒ \ z^2 \ - \ 6z \ + \ 9 \ = \ 0 \ ⇒ \ (z \ - \ 3)^2 \ = \ 0 \ ⇒ \ z \ = \ 3}$$

9) Solve for $$x$$: $$\sqrt{-9x \ + \ 27} \ = \ \sqrt{15x \ - \ 13}$$

$$\color{red}{\sqrt{-9x \ + \ 27} \ = \ \sqrt{15x \ - \ 13} \ ⇒ \ -9x \ + \ 27 \ = \ 15x \ - \ 13 \ ⇒ \ 40 \ = \ 24x \ ⇒ \ x \ = \ \frac{5}{3}}$$

10) Solve for $$n$$: $$\sqrt{-7n^2 \ + \ 26} \ = \ \sqrt{9n^2 \ - \ 18}$$

$$\color{red}{\sqrt{-7n^2 \ + \ 26} \ = \ \sqrt{9n^2 \ - \ 18} \ ⇒ \ -7n^2 \ + \ 26 \ = \ 9n^2 \ - \ 18 \ ⇒ \ 44 \ = \ 16n^2 \ ⇒ \ n^2 \ = \ \frac{11}{4}}$$ $$\color{red}{⇒ \ n \ = \ ± \frac{\sqrt{11}}{2}}$$

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