Lesson 102: Solving Quadratic Equations Using Square Roots

New Concept

Quadratic equations in the form $x^2 = a$ can be solved by taking the square root of both sides.

What’s next

Next, you’ll isolate the squared variable and use this principle to solve equations, finding both exact and approximate answers for various problems.

Property

For equations like $x^2 = a$, solve by taking the square root of both sides. This gives two solutions: $x = \pm\sqrt{a}$.

Explanation

Reversing a square usually gives two answers! Since both $5^2$ and $(-5)^2$ equal 25, the square root of 25 must be both 5 and -5. Don't forget the negative twin!

Examples

• $x^2 = 36 \implies x = \pm\sqrt{36} \implies x = \pm 6$
• $x^2 = -9 \implies$ No real solution exists.

How can a simple quadratic equation have two different answers? Let's see how taking a square root reveals both solutions.

Example Problem
Solve the equation $x^2 = 121$.

Step-by-Step

1. We start with the given equation where the variable is already squared and isolated.

2. Take the square root of both sides. Remember to include both the positive and negative roots on the right side.

3. Simplify the square roots to find the two possible values for $x$.

4. We can write the final solution using the plus-minus symbol.

5. Check both solutions:

Property

To solve an equation in the form $ax^2 + c = 0$, you must first isolate the $x^2$ term.

Explanation

It's a two-step puzzle! Before using your square root superpower, get $x^2$ all by itself. Use inverse operations to clear out the numbers tagging along with the variable.

Examples

• $3x^2 - 75 = 0 \implies 3x^2 = 75 \implies x^2 = 25 \implies x = \pm 5$
• $x^2 + 9 = 58 \implies x^2 = 49 \implies x = \pm 7$

Before you can take the square root, you have to clear away the other numbers. Let's practice isolating the squared variable first.

Example Problem
Solve the equation $5x^2 - 125 = 0$.

Step-by-Step

1. To solve for $x$, we first need to isolate the $x^2$ term.

2. Use the Addition Property of Equality to move the constant term to the other side.

3. Use the Division Property of Equality to isolate $x^2$.

4. Now that the equation is in the form $x^2 = a$, take the square root of both sides.

5. Simplify to find the final solutions.

6. Check both solutions:

Property

If $x^2$ equals a non-perfect square, the solutions are irrational. Simplify the radical first, then use a calculator to approximate.

Explanation

Not every number has a tidy square root. For messy ones like 50, the root is an endless decimal! First, simplify the radical, then use a calculator for a rounded estimate.

Examples

• $x^2 = 50 \implies x = \pm\sqrt{25 \cdot 2} = \pm 5\sqrt{2} \approx \pm 7.071$
• $x^2 = 15 \implies x = \pm\sqrt{15} \approx \pm 3.873$