Here, you will solve more complicated equations involving perfect squares.
There are two basic approaches you can use to solve an equation like [beautiful math coming... please be patient] $\,(3x+2)^2 = 16\,$:
To use this approach, you must:
| [beautiful math coming... please be patient] $\,(3x+2)^2 = 16\,$ | (original equation) |
| [beautiful math coming... please be patient] $\,(3x+2)^2 -16 = 0\,$ | (need $\,0\,$ on one side; subtract $\,16\,$ from both sides) |
| [beautiful math coming... please be patient] $(3x+2)^2 - 4^2 = 0$ | (rewrite, so it's clear you have a difference of squares) |
| [beautiful math coming... please be patient] $(3x+2+4)(3x+2-4) = 0$ | (factor the difference of squares) |
| [beautiful math coming... please be patient] $(3x+6)(3x-2) = 0$ | (simplify) |
| [beautiful math coming... please be patient] $3x+6 = 0\ \ \text{or}\ \ 3x-2 = 0$ | (use the zero factor law) |
| [beautiful math coming... please be patient] $3x = -6\ \ \text{or}\ \ 3x = 2$ | (solve the simpler equations) |
| [beautiful math coming... please be patient] $\displaystyle x = -2\ \ \text{or}\ \ x = \frac{2}{3}$ | (solve the simpler equations) |
| $(3(-2)+2)^2\ \ \overset{\text{?}}{=}\ \ 16$ | $(3(\frac{2}{3})+2)^2\ \ \overset{\text{?}}{=}\ \ 16$ |
| $(-6 + 2)^2 \ \ \overset{\text{?}}{=} \ \ 16$ | $(2 + 2)^2 \ \ \overset{\text{?}}{=} \ \ 16$ |
| $(-4)^2 \ \ \overset{\text{?}}{=} \ \ 16$ | $(4)^2 \ \ \overset{\text{?}}{=} \ \ 16$ |
| $16 = 16$ Check! | $16 = 16$ Check! |
The basic idea is that you're (correctly!) ‘undoing’ a square with the square root.
Notice that if
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$\,k\lt 0\,$,
then the equation $\,z^2 = k\,$ has no real number solutions.
For example, consider the equation
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$\,z^2 = -4\,$.
There is no real number which, when squared, gives $\,-4\,$.
To use this approach, you must:
| [beautiful math coming... please be patient] $(3x+2)^2 = 16$ | (original equation) |
| [beautiful math coming... please be patient] $3x + 2 = \pm\sqrt{16}$ | (check that $\,k\ge 0\,$; use the theorem) |
| [beautiful math coming... please be patient] $3x + 2 = \pm 4$ | (simplify: $\,\sqrt{16} = 4\,$) |
| [beautiful math coming... please be patient] $3x + 2 = 4\ \ \text{or}\ \ 3x + 2 = -4$ | (expand the ‘plus or minus’ shorthand) |
| [beautiful math coming... please be patient] $3x = 2\ \ \text{or}\ \ 3x = -6$ | (subtract $\,2\,$ from both sides of both equations) |
| [beautiful math coming... please be patient] $\displaystyle x = \frac{2}{3}\ \ \text{or}\ \ x = -2$ | (divide both sides of both equations by $\,3\,$) |
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 $\,(3x+2)^2 = 16\,$
is optionally accompanied by the
graph of $\,y = (3x+2)^2\,$ (the left side of the equation, dashed green)
and the graph of
$\,y = 16\,$ (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.