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]
$\,x^2 - 9 = 0\,$:
To use this approach, you must:
| [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) |
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:
| [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) |
Here are three slightly different approaches to solving the equation [beautiful math coming... please be patient] $\,16x^2 - 25 = 0\,$:
| [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) |
| [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) |
| [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) |
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.