audio read-through Solving Simple Equations Involving Perfect Squares

Here, you will solve simple equations involving perfect squares. There are two basic approaches you can use. They're both discussed thoroughly on this page.

Both approaches are illustrated next, using the equation $\,x^2 - 9 = 0\,$.

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:

$x^2 - 9 = 0$ original equation; check that $\,0\,$ is on one side of the equation
$(x+3)(x-3) = 0$ factor to get a product on the other side
$x+3 = 0\ \ \text{or}\ \ x-3 = 0$ use the zero factor law
$x=-3\ \ \text{or}\ \ x = 3$ solve the simpler equations

Approach #2 (Use the Following Theorem)

THEOREM solving equations involving perfect squares
For all real numbers $\,z\ $ and for $\,k\ge 0\,$: $$ \cssId{s27}{z^2 = k\ \ \ \text{is equivalent to}\ \ \ z=\pm\sqrt{k}} $$

Notice that if $\,k\lt 0\,,$ then the equation $\,z^2 = k\,$ has no real number solutions. For example, consider the equation $\,z^2 = -4\,.$ 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:

$x^2 - 9 = 0$ original equation
$x^2 = 9$ isolate a perfect square by adding $\,9\,$ to both sides
$x = \pm\sqrt{9}$ check that $\,k\ge 0\,$; use the theorem
$x = \pm 3$ rename: $\,\sqrt{9} = 3\,$
$x = 3\ \ \text{or}\ \ x = -3$ expand the ‘plus or minus’ shorthand, if desired

Examples

Here are three slightly different approaches to solving the equation $\,16x^2 - 25 = 0\,$:

First Approach: Use the Zero Factor Law
$16x^2 - 25 = 0$ original equation
$(4x)^2 - 5^2 = 0$ rewrite left-hand side as a difference of squares
$(4x + 5)(4x - 5) = 0$ factor the left-hand side
$4x + 5 = 0\ \ \text{or}\ \ 4x-5 = 0$ use the Zero Factor Law
$4x = -5\ \ \text{or}\ \ 4x = 5$ solve the simpler equations
$\displaystyle x = -\frac{5}{4}\ \ \text{or}\ \ x = \frac{5}{4}$ solve the simpler equations
Second Approach: Use the theorem; isolate $\,x^2\,$
$16x^2 - 25 = 0$ original equation
$16x^2 = 25$ add $\,25\,$ to both sides
$\displaystyle x^2 = \frac{25}{16}$ divide both sides by $\,16\,$:  now, $\,x^2\,$ is isolated
$\displaystyle x = \pm\sqrt{\frac{25}{16}}$ use the theorem
$\displaystyle x = \pm\frac{5}{4}$ rename: $\,\sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4}$
$\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = -\frac{5}{4}$ expand the ‘plus or minus’ shorthand, if desired
Third Approach: Use the theorem; isolate $\,(4x)^2\ $
$16x^2 - 25 = 0$ original equation
$16x^2 = 25$ add $\,25\,$ to both sides
$(4x)^2 = 25$ rename left-hand side as a perfect square
$4x = \pm\sqrt{25}$ use the theorem
$4x = \pm 5$ rename: $\,\sqrt{25} = 5\,$
$\displaystyle x = \frac{\pm 5}{4}$ divide both sides by $\,4\,$
$\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = -\frac{5}{4}$ expand the ‘plus or minus’ shorthand, if desired

Concept Practice

For more advanced students, a graph is available. 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 to toggle the graph.


Solve: