Difference between revisions of "2018 AMC 10A Problems/Problem 10"
(→Solution 3) |
Rockmanex3 (talk | contribs) (Problem Type) |
||
Line 64: | Line 64: | ||
</asy> | </asy> | ||
− | {{AMC10 box|year=2018|ab=A|num-b=9| | + | ==See Also== |
+ | |||
+ | {{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Revision as of 15:25, 18 June 2018
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 = 3, . - cookiemonster2004
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 | |
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 |