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 Equations using 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

$$x^2 = k$$ $$x = \sqrt{k} \hspace{.5em} or \hspace{.5em} x = -\sqrt{k}$$ 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: $$x = 5 \hspace{.5em} or \hspace{.5em} x = -5$$ 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: $$x = 8 \hspace{.5em} or \hspace{.5em} x = -\frac{26}{3}$$ 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: $$x = \frac{3}{5} \hspace{.5em} or \hspace{.5em} -\frac{9}{5}$$