Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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Substituting into the equation <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>, we get <math>\sqrt{49-\frac{75}{4}}+\sqrt{25-\frac{75}{4}}</math>. Immediately, we simplify into <math>\sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}}</math>. The two numbers inside the square roots are simplified to be <math>\frac{11}{2}</math> and <math>\frac{5}{2}</math>, so you add them up: <math>\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}</math> | Substituting into the equation <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>, we get <math>\sqrt{49-\frac{75}{4}}+\sqrt{25-\frac{75}{4}}</math>. Immediately, we simplify into <math>\sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}}</math>. The two numbers inside the square roots are simplified to be <math>\frac{11}{2}</math> and <math>\frac{5}{2}</math>, so you add them up: <math>\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}</math> | ||
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{{AMC10 box|year=2018|ab=A|num-b=9|after=Problem 11}} | {{AMC10 box|year=2018|ab=A|num-b=9|after=Problem 11}} |
Revision as of 13:47, 11 February 2018
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 = 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 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:
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 |