Lesson 9.6: Solve Equations with Square Roots

New Concept

This lesson introduces radical equations, where the variable is inside a square root. You'll learn to solve them by isolating the radical and squaring both sides. This technique unlocks solutions for real-world problems involving geometry and physics formulas.

What’s next

Next, you'll master this concept through interactive examples. Then, you'll apply your skills to solve challenge problems using real-world formulas.

Property

An equation in which the variable is in the radicand of a square root is called a radical equation.
When we solve radical equations by squaring both sides we may get an algebraic solution that would make $\sqrt{a}$ negative.
This algebraic solution would not be a solution to the original radical equation; it is an extraneous solution.

Examples

• For the equation $\sqrt{x+6} = x$, is $x=3$ a solution? Yes, because $\sqrt{3+6} = \sqrt{9} = 3$.

• For the same equation, $\sqrt{x+6} = x$, is $x=-2$ a solution? No, because $\sqrt{-2+6} = \sqrt{4} = 2$, and $2 \ne -2$. So, $x=-2$ is an extraneous solution.

Property

To solve a radical equation:

1. Isolate the radical on one side of the equation.
2. Square both sides of the equation.
3. Solve the new equation.
4. Check the answer.

This strategy is based on the property that for $a \ge 0$, $(\sqrt{a})^2 = a$.

Examples

• To solve $\sqrt{3x-2} = 5$, square both sides: $(\sqrt{3x-2})^2 = 5^2$, which gives $3x-2 = 25$. Solving for $x$ gives $3x = 27$, so $x=9$.

• To solve $\sqrt{4n-3} - 7 = 0$, first isolate the radical: $\sqrt{4n-3} = 7$. Square both sides: $4n-3 = 49$. This gives $4n=52$, so $n=13$.

Property

When squaring a binomial, use the binomial squares formulas. Don't forget the middle term!

Examples

• Solve $\sqrt{x-2} + 2 = x$. First, isolate the radical: $\sqrt{x-2} = x-2$. Squaring both sides gives $x-2 = (x-2)^2 = x^2 - 4x + 4$. The solutions are $x=2$ and $x=3$.

• Solve $\sqrt{k+7} = k-5$. Squaring both sides gives $k+7 = (k-5)^2 = k^2 - 10k + 25$. This simplifies to $0 = k^2 - 11k + 18$, which factors to $(k-2)(k-9)$. Only $k=9$ is a valid solution.

Property

Area, $A = s^2$
Length of a side, $s = \sqrt{A}$
We can use the formula $s = \sqrt{A}$ to find the length of a side of a square for a given area.

Examples

• A square garden has an area of 150 square feet. The length of each side is $s = \sqrt{150} \approx 12.2$ feet.

• A square photo has an area of 81 square inches. The length of each side is $s = \sqrt{81} = 9$ inches.

Property

On Earth, if an object is dropped from a height of $h$ feet, the time in seconds it will take to reach the ground is found by using the formula,

Examples

• A ball is dropped from a 144-foot-tall building. The time it takes to hit the ground is $t = \frac{\sqrt{144}}{4} = \frac{12}{4} = 3$ seconds.

• A phone is dropped from a bridge 256 feet above a river. The time it takes to reach the water is $t = \frac{\sqrt{256}}{4} = \frac{16}{4} = 4$ seconds.

Property

If the length of the skid marks is $d$ feet, then the speed, $s$, of the car before the brakes were applied can be found by using the formula,

Examples

• Skid marks at an accident scene measure 150 feet. The car's estimated speed was $s = \sqrt{24(150)} = \sqrt{3600} = 60$ miles per hour.

• After an emergency stop, a car's skid marks measured 96 feet. Its speed was $s = \sqrt{24(96)} = \sqrt{2304} = 48$ miles per hour.