Word Problems Involving Perfect Squares
Need some practice solving equations involving perfect squares first?
- Solving Simple Equations involving Perfect Squares
- Solving More Complicated Equations involving Perfect Squares
Here, you will solve word problems that result in equations involving perfect squares.
Examples
Question:
I'm thinking of a number.
The square of $\,3\,$ times this number is $\,25\,.$
What number(s) could I be thinking of?
Solution:
$(3x)^2 = 25$
$3x = \pm 5$
$3x = 5\ \ \text{or}\ \ 3x = -5$
$x = \frac{5}{3}\ \ \text{or}\ \ x = -\frac{5}{3}$
$(3x)^2 = 25$
$3x = \pm 5$
$3x = 5\ \ \text{or}\ \ 3x = -5$
$x = \frac{5}{3}\ \ \text{or}\ \ x = -\frac{5}{3}$
Question:
I'm thinking of a number.
When I take one less than three times this number,
and then square the result,
I end up with the number $\,25\,.$
What number(s) could I be thinking of?
Solution:
$(3x-1)^2 = 25$
$3x-1 = \pm 5$
$3x-1 = 5\ \ \text{or}\ \ 3x-1 = -5$
$3x = 6\ \ \text{or}\ \ 3x = -4$
$x = 2\ \ \text{or}\ \ x = -\frac{4}{3}$
$(3x-1)^2 = 25$
$3x-1 = \pm 5$
$3x-1 = 5\ \ \text{or}\ \ 3x-1 = -5$
$3x = 6\ \ \text{or}\ \ 3x = -4$
$x = 2\ \ \text{or}\ \ x = -\frac{4}{3}$
Question:
I'm thinking of a negative number.
When I take the sum of this number and $\,2\,,$
and then square the result,
I end up with the number $\,9\,.$
What number am I thinking of?
Solution:
$(x+2)^2 = 9$
$x+2 = \pm 3$
$x+2 = 3\ \ \text{or}\ \ x+2 = -3$
$x = 1\ \ \text{or}\ \ x = -5$
Since the number being thought of is negative, the answer is $\,-5\,.$
$(x+2)^2 = 9$
$x+2 = \pm 3$
$x+2 = 3\ \ \text{or}\ \ x+2 = -3$
$x = 1\ \ \text{or}\ \ x = -5$
Since the number being thought of is negative, the answer is $\,-5\,.$
Question:
I'm thinking of a positive number.
When I take the difference of this number and $\,3\,,$
and then square the result,
I end up with the number
$\,16\,.$
What number am I thinking of?
Solution:
$(x-3)^2 = 16$
$x-3 = \pm 4$
$x-3 = 4\ \ \text{or}\ \ x-3 = -4$
$x = 7\ \ \text{or}\ \ x = -1$
Since the number being thought of is positive, the answer is $\,7\,.$
$(x-3)^2 = 16$
$x-3 = \pm 4$
$x-3 = 4\ \ \text{or}\ \ x-3 = -4$
$x = 7\ \ \text{or}\ \ x = -1$
Since the number being thought of is positive, the answer is $\,7\,.$
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Concept Practice
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