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
[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:

- 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 [beautiful math coming... please be patient] $\,a\,$ and $\,b\,$, $\,ab = 0\,$ is equivalent to $\,(a = 0\ \ \text{or}\ \ b = 0)\,$.

[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:

- Isolate a perfect square on one side of the equation.
- Check that you have a
*nonnegative*number on the other side. - Use the theorem.

[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

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.