Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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Revision as of 22:59, 9 February 2018
Contents
[hide]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 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 |