Find practice problems and solutions for these concepts at Solving Radical Equations Practice Problems.
In this lesson, you will solve equations that contain radicals. You will use the skills you have acquired throughout this book to solve these equations. You may also need to simplify radicals to solve the equations in this lesson.
What Is a Radical Equation?
What is a radical equation? An equation is not considered a radical equation unless the radicand contains a variable, like √x = 3. You know that squaring something is the opposite of taking the square root. To solve a radical equation, you square both sides.
Example: √x = 3
Square both sides. (√x)^{2} = 3^{2}
Simplify. √x · √x = 3 · 3
Multiply. √x · x = 9
Take the square root of x^{2}. x = 9
Solving an equation like x^{2} = 25 requires a little extra thought. Plug in x = 5 and you see that 5^{2} = 25. This means that x = 5 is a solution to x^{2} = 25. However, if you plug in x = –5, you see that (–5)^{2} = (–5) · (–5) = 25. This means that x = –5 is also a solution to x^{2} = 25, so the equation has two solutions: x = 5 and x = –5. This happens so often that there is a special symbol ± that means plus or minus. You say that x = ±5 is the solution to x^{2} = 25.
Remember that every quadratic equation has two solutions.
Example: x ^{2} = 24 | |
Take the square root of both sides. | √x2 = √24 |
The answer could be + or –. | x = ±√24 |
Factor out perfect squares. | x = ±√4 · 6 |
Simplify. | x = ±2√6 |
Solving Complex Radical Equations
Now that you know what a radical equation is, use what you have learned about solving equations to solve radical equations that require more than one step.
TipBefore squaring both sides of an equation, get the radical on a side by itself. |
Example: √x + 1 = 5 | |
Subtract 1 from both sides of the equation. | √x + 1 – 1 = 5 – 1 |
Simplify. | √x = 4 |
Square both sides of the equation. | (√x)^{2} = 4^{2} |
Simplify. | x = 16 |
Example: 3√x = 15 | |
Divide both sides of the equation by 3. | |
Simplify. | √x = 5 |
Square both sides of the equation. | (√x)^{2} = 5^{2} |
Simplify. | x = 25 |
Example: 2√x + 2 = 10 | |
Divide both sides of the equation by 2. | |
Simplify. | √x + 2 = 5 |
Square both sides of the equation. | (√x + 2)^{2} = 5^{2} |
Simplify. | x + 2 = 25 |
Subtract 2 from both sides of the equation. | x + 2 – 2 = 25 – 2 |
Simplify. | x = 23 |
Example: 2√x + 2 = 18 | |
Subtract 2 from both sides of the equation. | 2√x + 2 – 2 = 18 – 2 |
Simplify. | 2√x = 16 |
Divide both sides of the equation by 2. | |
Simplify | √x = 8 |
Square both sides of the equation. | (√x)^{2} = 8^{2} |
Simplify. | x = 64 |
TipWhen you multiply a radical by itself, the radical sign disappears. Example: √x · √x = x and √x + 3 · √x + 3 = x + 3 |
Skill Building until Next TimeThe higher you climb, the farther you can see. There is a formula for this: V = 3.5√h, where h is your height, in meters, above the ground, and, V is the distance, in kilometers, that you can see. The next tie you are in a building with more than one story, look out a window on different floors of the building. Do you notice a difference in how far you can see? Calculate how far the distance would be. Use 3 meters for each floor in the building. If you were on the ninth floor, the height would be 9 · 3 = 27 meters. |
Find practice problems and solutions for these concepts at Solving Radical Equations Practice Problems.
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