Equations with Radicals

Main idea

Get the root alone on one side, then square (or cube) both sides to get rid of it. Squaring can sneak in fake solutions, so every answer has to be checked in the original equation.

\sqrt{f(x)} = g(x) \implies f(x) = {\htmlData{sym=0}{g(x)^2}} \quad\text{(then check!)}

What the symbols mean

$g(x)^2$
Squared side
Squaring loses the sign information, which is why extra solutions can appear.

Explanation

A radical equation is simply one where the unknown is stuck under a root sign, like $\sqrt{x + 2} = x$ . The way to free it is to undo the root: squaring undoes a square root, cubing undoes a cube root. The formula up top says just that, with $\sqrt{f(x)}$ (the root side), $g(x)$ (the other side) and $g(x)^2$ (what you get after squaring), plus a reminder to check. That last word is the important part of this lesson.

Why isolate first? Squaring a sum doesn't get rid of a root. $(\sqrt x + 1)^2 = x + 2\sqrt x + 1$ still has $\sqrt x$ in it. Squaring only clears the root cleanly when the root is alone on its side.

Why check? Here's the catch. Squaring is not a reversible move the way adding or dividing is. $x = 3$ and $x^2 = 9$ aren't the same statement: the second also lets in $x = -3$ . In the same way, squaring $\sqrt{x + 2} = x$ forgets that a square root can never be negative, so it can produce answers that would need $\sqrt{\ }$ to equal a negative number. Those are extraneous, and plugging back into the original is how you catch them.

Two radicals. Isolate one, square, simplify, and if a root is still there, isolate it and square again.

See it

Solving √(x + 2) = x − c by squaring gives a quadratic with two roots. But squaring also lets in the dashed lower branch, y = −√(x + 2), which was never part of the original equation. Slide the line and notice one root always lands on that fake branch. It's extraneous, and that's why answers have to be checked.

Worked examples

Example 1

Solve $\sqrt{x + 2} = x$ .

1
The root is already alone, so square both sides: $x + 2 = x^2$ . That gives $x^2 - x - 2 = 0$ , which factors to $x = 2$ or $x = -1$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Solve $\sqrt{x + 2} = x$ .

Problem 2+5 XP

Solve $\sqrt{2x - 1} = 3$ .

Problem 3+5 XP

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

Problem 4+5 XP

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

Problem 5+5 XP

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

Problem 6+5 XP

A pendulum's period is $T = 2\pi\sqrt{\frac{L}{9.8}}$ seconds, with $L$ in meters. What length gives a period of exactly 2 seconds?

Problem 7+5 XP

Solve $\sqrt{3x + 1} = x - 1$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $\sqrt{x + 5} - \sqrt{x - 3} = 2$ .

Where it's used

Pendulum lengths from periods, distances from areas and speeds from braking distances all require solving radical equations.

How Fast a Tsunami TravelsAlgebra · Physics

Equations Reducible to Quadratic Form Linear Inequalities

Equations with Radicals

Main idea

Get the root alone on one side, then square (or cube) both sides to get rid of it. Squaring can sneak in fake solutions, so every answer has to be checked in the original equation.

\sqrt{f(x)} = g(x) \implies f(x) = {\htmlData{sym=0}{g(x)^2}} \quad\text{(then check!)}

What the symbols mean

$g(x)^2$
Squared side
Squaring loses the sign information, which is why extra solutions can appear.

Explanation

Two radicals. Isolate one, square, simplify, and if a root is still there, isolate it and square again.

See it

Worked examples

Example 1

Solve $\sqrt{x + 2} = x$ .

1
The root is already alone, so square both sides: $x + 2 = x^2$ . That gives $x^2 - x - 2 = 0$ , which factors to $x = 2$ or $x = -1$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Solve $\sqrt{x + 2} = x$ .

Problem 2+5 XP

Solve $\sqrt{2x - 1} = 3$ .

Problem 3+5 XP

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

Problem 4+5 XP

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

Problem 5+5 XP

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

Problem 6+5 XP

A pendulum's period is $T = 2\pi\sqrt{\frac{L}{9.8}}$ seconds, with $L$ in meters. What length gives a period of exactly 2 seconds?

Problem 7+5 XP

Solve $\sqrt{3x + 1} = x - 1$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $\sqrt{x + 5} - \sqrt{x - 3} = 2$ .

Where it's used

Pendulum lengths from periods, distances from areas and speeds from braking distances all require solving radical equations.

How Fast a Tsunami TravelsAlgebra · Physics