SOLVING SIMPLE EQUATIONS INVOLVING PERFECT SQUARES
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Here, you will solve simple equations involving perfect squares.
There are two basic approaches you can use to solve an equation like [beautiful math coming... please be patient] $\,x^2 - 9 = 0\,$:

Approach #1 (factor and use the Zero Factor Law)

To use this approach, you must:

To use this approach, you would write the following list of equivalent sentences:
[beautiful math coming... please be patient] $x^2 - 9 = 0$ (original equation; check that $\,0\,$ is on one side of the equation)
[beautiful math coming... please be patient] $(x+3)(x-3) = 0$ (factor to get a product on the other side)
[beautiful math coming... please be patient] $x+3 = 0\ \ \text{or}\ \ x-3 = 0$ (use the zero factor law)
[beautiful math coming... please be patient] $x=-3\ \ \text{or}\ \ x = 3$ (solve the simpler equations)

Approach #2 (use the following theorem)
THEOREM solving equations involving perfect squares
For all real numbers $\,z\ $ and for $\,k\ge 0\,$: [beautiful math coming... please be patient] $$ z^2 = k\ \ \ \text{is equivalent to}\ \ \ z=\pm\sqrt{k} $$

Notice that if [beautiful math coming... please be patient] $\,k\lt 0\,$, then the equation $\,z^2 = k\,$ has no real number solutions.
For example, consider the equation [beautiful math coming... please be patient] $\,z^2 = -4\,$.
There is no real number which, when squared, gives $\,-4\,$.

To use this approach, you must:

To use this approach, you would write the following list of equivalent sentences:
[beautiful math coming... please be patient] $x^2 - 9 = 0$ (original equation)
[beautiful math coming... please be patient] $x^2 = 9$ (isolate a perfect square by adding $\,9\,$ to both sides)
[beautiful math coming... please be patient] $x = \pm\sqrt{9}$ (check that $\,k\ge 0\,$; use the theorem)
[beautiful math coming... please be patient] $x = \pm 3$ (rename: $\,\sqrt{9} = 3\,$)
[beautiful math coming... please be patient] $x = 3\ \ \text{or}\ \ x = -3$ (expand the ‘plus or minus’ shorthand, if desired)

EXAMPLES:

Here are three slightly different approaches to solving the equation [beautiful math coming... please be patient] $\,16x^2 - 25 = 0\,$:

First Approach (use the Zero Factor Law)

[beautiful math coming... please be patient] $16x^2 - 25 = 0$ (original equation)
[beautiful math coming... please be patient] $(4x)^2 - 5^2 = 0$ (rewrite left-hand side as a difference of squares)
[beautiful math coming... please be patient] $(4x + 5)(4x - 5) = 0$ (factor the left-hand side)
[beautiful math coming... please be patient] $4x + 5 = 0\ \ \text{or}\ \ 4x-5 = 0$ (use the Zero Factor Law)
[beautiful math coming... please be patient] $4x = -5\ \ \text{or}\ \ 4x = 5$ (solve the simpler equations)
[beautiful math coming... please be patient] $\displaystyle x = -\frac{5}{4}\ \ \text{or}\ \ x = \frac{5}{4}$ (solve the simpler equations)
Second Approach (use the theorem; isolate $\,x^2\,$)

[beautiful math coming... please be patient] $16x^2 - 25 = 0$ (original equation)
[beautiful math coming... please be patient] $16x^2 = 25$ (add $\,25\,$ to both sides)
[beautiful math coming... please be patient] $\displaystyle x^2 = \frac{25}{16}$ (divide both sides by $\,16\,$:   now, $x^2$ is isolated)
[beautiful math coming... please be patient] $\displaystyle x = \pm\sqrt{\frac{25}{16}}$ (use the theorem)
[beautiful math coming... please be patient] $\displaystyle x = \pm\frac{5}{4}$ (rename: [beautiful math coming... please be patient] $\,\sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4}$)
[beautiful math coming... please be patient] $\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = -\frac{5}{4}$ (expand the ‘plus or minus’ shorthand, if desired)
Third Approach (use the theorem; isolate $\,(4x)^2\ $)

[beautiful math coming... please be patient] $16x^2 - 25 = 0$ (original equation)
[beautiful math coming... please be patient] $16x^2 = 25$ (add $\,25\,$ to both sides)
[beautiful math coming... please be patient] $(4x)^2 = 25$ (rename left-hand side as a perfect square)
[beautiful math coming... please be patient] $4x = \pm\sqrt{25}$ (use the theorem)
[beautiful math coming... please be patient] $4x = \pm 5$ (rename: $\,\sqrt{25} = 5\,$)
[beautiful math coming... please be patient] $\displaystyle x = \frac{\pm 5}{4}$ (divide both sides by $\,4\,$)
[beautiful math coming... please be patient] $\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = -\frac{5}{4}$ (expand the ‘plus or minus’ shorthand, if desired)
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Solving More Complicated Equations
Involving Perfect Squares

 
 

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.

For more advanced students, a graph is displayed.
For example, the equation $\,x^2 - 9 = 0\,$
is optionally accompanied by the graph of $\,y = x^2 - 9\,$ (the left side of the equation, dashed green)
and the graph of $\,y = 0\,$ (the right side of the equation, solid purple).
In this example, you are finding the values of $\,x\,$ where the green graph intersects the purple graph.
Click the “show/hide graph” button if you prefer not to see the graph.

Solve: