Difference between revisions of "2019 AMC 8 Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
| − | We know that to get 16, you can square 4 or -4. Thus, x | + | We know that to get <math>16</math>, you can square <math>4</math> or <math>-4</math>. Thus, <math>x^2 - 5</math> can have <math>2</math> possibilities. <math>x^2</math> is either <math>9</math> or <math>1</math>, leaving <math>x</math> with possiblities <math>3,-3, 1,</math> and <math>-1,</math> so <math>\boxed{(D)}</math>. |
==See Also== | ==See Also== | ||
Revision as of 19:41, 20 November 2019
Problem 20
How many different real numbers 
satisfy the equation ![\[(x^{2}-5)^{2}=16?\]](http://latex.artofproblemsolving.com/d/1/8/d188c4697f3f7cf1889796a037bd60b6dae3b29e.png)
Solution 1
We know that to get 
, you can square 
or 
. Thus, 
can have 
possibilities. 
is either 
or 
, leaving with possiblities 
and 
so 
.
See Also
| 2019 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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