Lesson Objectives
• Demonstrate an understanding of how to solve a quadratic equation by factoring
• Learn about the square root property
• Learn how to solve equations of the form: x2 = k
• Learn how to solve equations of the form: (ax + b)2 = k

## How to Solve Quadratic Equations with the Square Root Property

At this point, we should be comfortable with solving a quadratic equation using factoring and the zero-product property. As an example: $$x^2 + 2x - 3=0$$ First, we will factor the left side of the equation: $$(x - 1)(x + 3)=0$$ Now we can use the zero-product property, this means we will set each factor equal to zero and solve: $$x - 1=0$$ $$x=1$$ $$x + 3=0$$ $$x=-3$$ We can see our two solutions for the equation:
x = 1 or x = -3
We won't be able to solve every quadratic equation using factoring. Over the next few lessons, we will develop strategies to quickly solve any quadratic equation whether it can be factored or not.

### Square Root Property of Equations

$$x^2=k$$ $$x=\sqrt{k}\hspace{.5em}\text{or} \hspace{.5em}x=-\sqrt{k}$$ This can be written using a shorthand notation: $$x=\pm \sqrt{k}$$ The plus on top of the minus means plus or minus. This allows us to write two solutions where only the sign differs in a more compact format. Let's suppose we saw something such as: $$x^2=25$$ We could solve this using factoring: $$x^2 - 25=0$$ $$(x + 5)(x - 5)=0$$ $$x + 5=0$$ $$x=-5$$ $$x - 5=0$$ $$x=5$$ Our solution: $$x=5 \hspace{.5em}\text{or} \hspace{.5em}x=-5$$ Using our shorthand notation: $$x=\pm 5$$ We can also solve this type of equation using our square root property: $$x^2=25$$ $$\sqrt{x^2}=\pm \sqrt{25}$$ $$x=\pm 5$$ Let's look at a few examples.
Example 1: Solve each equation. $$x^2=81$$ We can use our square root property to solve the equation: $$\sqrt{x^2}=\pm \sqrt{81}$$ $$x=\pm 9$$ Example 2: Solve each equation. $$4x^2=676$$ Let's begin by dividing each side of the equation by 4. This will place our equation in the format of:
x2 = k
$$\require{cancel}\frac{\cancel{4}x^2}{\cancel{4}}=\frac{169 \cancel{676}}{\cancel{4}}$$ $$x^2=169$$ We can use our square root property to solve the equation: $$\sqrt{x^2}=\pm \sqrt{169}$$ $$x=\pm 13$$ Example 3: Solve each equation. $$49x^2 + 17=81$$ We want the equation in the format of:
x2 = k
We can first subtract 17 away from each side:
$$49x^2=64$$ Now we will divide each side by 49: $$\frac{\cancel{49}x^2}{\cancel{49}}=\frac{64}{49}$$ $$x^2=\frac{64}{49}$$ We can use our square root property to solve the equation: $$\sqrt{x^2}=\pm \sqrt{\frac{64}{49}}$$ $$x=\pm \frac{8}{7}$$ Additionally, we can use this rule when we have a binomial squared. Let's look at a few examples.
Example 4: Solve each equation. $$(3x + 1)^2=625$$ We can use our square root property to solve the equation: $$\sqrt{(3x + 1)^2}=\pm \sqrt{625}$$ $$3x + 1=\pm 25$$ We have to solve two equations here: $$3x + 1=25$$ $$3x=24$$ $$x=8$$ $$3x + 1=-25$$ $$3x=-26$$ $$x=-\frac{26}{3}$$ Our solutions: $$x=8 \hspace{.5em}\text{or} \hspace{.5em}x=-\frac{26}{3}$$ Example 5: Solve each equation. $$(5x + 3)^2=36$$ We can use our square root property to solve the equation: $$\sqrt{(5x + 3)^2}=\pm \sqrt{36}$$ $$5x + 3=\pm 6$$ We have to solve two equations here: $$5x + 3=6$$ $$5x=3$$ $$x=\frac{3}{5}$$ $$5x + 3=-6$$ $$5x=-9$$ $$x=-\frac{9}{5}$$ Our solutions: $$x=\frac{3}{5}\hspace{.5em}\text{or} \hspace{.5em}-\frac{9}{5}$$

#### Skills Check:

Example #1

Solve each equation. $$3x^2 - 7=77$$

Please choose the best answer.

A
$$x=\pm 5$$
B
$$x=\pm 3$$
C
$$x=\pm \sqrt{7}$$
D
$$x=\pm 2\sqrt{7}$$
E
$$x=\pm \frac{\sqrt{7}}{2}$$

Example #2

Solve each equation. $$9x^2 - 5=-18$$

Please choose the best answer.

A
$$x=\pm \frac{i \sqrt{13}}{3}$$
B
$$x=\pm 2$$
C
$$x=\pm 3i$$
D
$$x=\pm 2i\sqrt{13}$$
E
$$x=\pm i\sqrt{13}$$

Example #3

Solve each equation. $$(2x - 3)^2=16$$

Please choose the best answer.

A
$$x=\pm \frac{i}{2}$$
B
$$x=-\frac{1}{2}, \frac{7}{2}$$
C
$$x=\pm 3i$$
D
$$x=\pm \frac{5i}{2}$$
E
$$x=-3, 7$$

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