Difference between revisions of "2019 AMC 8 Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
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| + | We know that to get 16, you can square 4 or -4. Thus, x squared - 5 can have 2 possibilities. x squared is either 9 or 1, leaving x with possiblities 3,-3, 1, and -1. | ||
==See Also== | ==See Also== | ||
Revision as of 19:21, 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 16, you can square 4 or -4. Thus, x squared - 5 can have 2 possibilities. x squared is either 9 or 1, leaving x with possiblities 3,-3, 1, and -1.
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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