2018 AMC 10A Problems/Problem 10
Contents
Problem
Suppose that real number satisfies
What is the value of
?
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 . - cookiemonster2010
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 by 6 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:
.
Solution 4 (Geometric Interpretation)
Draw a right triangle with a hypotenuse
of length
and leg
of length
. Draw
on
such that
. Note that
and
. Thus, from the given equation,
. Using Law of Cosines on triangle
, we see that
so
. Since
is a
triangle,
and
. Finally,
.
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AMC 10 Problems and Solutions |