SOLVING SIMPLE EQUATIONS INVOLVING PERFECT SQUARES

Here, you will solve simple equations involving perfect squares.
There are two basic approaches you can use to solve an equation like $\,x^2 - 9 = 0\,$:

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

To use this approach, you must:

• Check that you have $\,0\,$ on one side of the equation.
• Factor to get a product on the other side of the equation.
• Use the Zero Factor Law:
For all real numbers $\,a\,$ and $\,b\,$, $\,ab = 0\,$ is equivalent to $\,(a = 0\ \ \text{or}\ \ b = 0)\,$.
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 $\,k\ge 0\,$: $$z^2 = k\ \ \ \text{is equivalent to}\ \ \ z=\pm\sqrt{k}$$

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:

• Isolate a perfect square on one side of the equation.
• Check that you have a nonnegative number on the other side.
• Use the theorem.
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)
Master the ideas from this section

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

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.