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: $$x = \frac{3}{5} \hspace{.5em} or \hspace{.5em} -\frac{9}{5}$$