Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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Revision as of 22:59, 9 February 2018
Solutions
Solution 1
In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The 
terms cancel nicely. 
Given that 
= 3, 
Solution by PancakeMonster2004, explanations added by a1b2.
Solution 2
Let 
, and let 
. Then 
. Substituting, we get 
. Rearranging, we get 
. Squaring both sides and solving, we get 
and 
. Adding, we get that the answer is 
Solution 3
Put the equations to one side. 
can be changed into 
.
We can square both sides, getting us 
That simplifies out to 
Dividing both sides gets us 
.
Following that, we can square both sides again, resulting in the equation 
. Simplifying that, we get 
.
Substituting into the equation 
, we get 
. Immediately, we simplify into 
. The two numbers inside the square roots are simplified to be 
and 
, so you add them up: 
~kevinmathz
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AMC 10 Problems and Solutions | ||
