Lesson 10.1: Solve Quadratic Equations Using the Square Root Property

New Concept

The Square Root Property provides a direct way to solve quadratic equations in the form $ax^2=k$ or $a(x-h)^2=k$. By isolating the squared term, you can find the solutions by taking the square root of both sides.

What’s next

Next, you’ll work through guided examples showing how to apply this property. Then, you'll tackle a series of practice problems to build your skills.

Property

Quadratic equations are equations of the form $ax^2 + bx + c = 0$, where $a \neq 0$.
They differ from linear equations by including a term with the variable raised to the second power.
We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

Examples

• The equation $x^2 = 100$ is a quadratic equation where the $x$ is squared.
• An equation like $5y^2 + 10y - 15 = 0$ is a quadratic equation in standard form.
• The equation $2z^2 = 50$ is a quadratic equation of the form $ax^2 = k$.

Explanation

Think of a quadratic equation as a 'squared' equation. Unlike simple linear equations, you can't isolate the variable with basic arithmetic. It requires special tools like factoring or using square roots to find the solution(s).

Property

If $x^2 = k$, and $k \ge 0$, then $x = \sqrt{k}$ or $x = -\sqrt{k}$. The solution can also be written as $x = \pm \sqrt{k}$.

Examples

• To solve $x^2 = 81$, we apply the Square Root Property to get $x = \pm\sqrt{81}$, which gives the solutions $x = 9$ and $x = -9$.
• For the equation $y^2 = 15$, the solutions are $y = \pm\sqrt{15}$, which we leave in radical form as $y = \sqrt{15}$ and $y = -\sqrt{15}$.
• If $z^2 = -9$, there is no real solution because the square root of a negative number is not a real number.

Explanation

When a variable squared equals a number, the variable itself can be either the positive or negative square root of that number. This is because squaring a negative value results in a positive value, creating two possible answers.

Property

To solve a quadratic equation using the Square Root Property:

1. Isolate the quadratic term and make its coefficient one.
2. Use the Square Root Property.
3. Simplify the radical.
4. Check the solutions.

Examples

• To solve $4x^2 = 100$, first divide by 4 to get $x^2 = 25$. Then, using the Square Root Property, $x = \pm\sqrt{25}$, so $x = 5$ and $x = -5$.
• Solve $3y^2 - 54 = 0$. First, add 54 to both sides to get $3y^2 = 54$. Divide by 3 to get $y^2 = 18$. The solution is $y = \pm\sqrt{18} = \pm 3\sqrt{2}$.
• Solve $3c^2 = 75$. Divide by 3 to get $c^2=25$. Using the property, $c = \pm\sqrt{25}$, so $c=5, c=-5$.

Explanation

Before you can apply the Square Root Property, the $x^2$ term must be by itself. To achieve this, divide both sides of the equation by the coefficient of $x^2$, and then proceed with taking the square root.

Property

We can use the Square Root Property to solve an equation like $(x-3)^2 = 16$, too.
We will treat the whole binomial, $(x-3)$, as the quadratic term.
To solve, first isolate the binomial term, then use the Square Root Property, and finally solve for the variable.

Examples

• To solve $(x-5)^2 = 49$, take the square root of both sides: $x-5 = \pm\sqrt{49}$, so $x-5 = \pm 7$. This leads to two solutions: $x = 12$ and $x = -2$.
• Solve $2(y+3)^2 = 50$. First, divide by 2 to get $(y+3)^2 = 25$. Then $y+3 = \pm\sqrt{25}$, so $y+3 = \pm 5$. The solutions are $y = 2$ and $y = -8$.
• Solve $(z-6)^2 = 12$. Take the square root: $z-6 = \pm\sqrt{12} = \pm 2\sqrt{3}$. The final solutions are $z = 6 \pm 2\sqrt{3}$.

Explanation

This may look more complex, but the strategy is the same. Treat the entire expression in the parentheses as a single squared item. Isolate it, take the square root of both sides, and then solve the remaining simple equation.

Property

Some equations are not initially in the form $a(x-h)^2 = k$, but the left side is a perfect square trinomial.
We will factor it first to put it in the form we need.

Examples

• Solve $x^2 - 14x + 49 = 25$. The left side factors to $(x-7)^2$. The equation becomes $(x-7)^2 = 25$, so $x-7 = \pm 5$. The solutions are $x = 12$ and $x = 2$.
• Solve $y^2 + 6y + 9 = 7$. Factor the left side to get $(y+3)^2 = 7$. Using the Square Root Property, $y+3 = \pm\sqrt{7}$, so the solution is $y = -3 \pm\sqrt{7}$.
• Solve $9z^2 + 12z + 4 = 16$. The left side factors to $(3z+2)^2$. The equation is $(3z+2)^2 = 16$, so $3z+2 = \pm 4$. The two solutions are $z = \frac{2}{3}$ and $z = -2$.

Explanation

If you spot a special trinomial like $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2$, you can factor it into $(a+b)^2$ or $(a-b)^2$. This transforms the problem into a familiar form that you can solve with the Square Root Property.