Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The <math>x^2</math> terms cancel nicely. <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24</math> | In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The <math>x^2</math> terms cancel nicely. <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24</math> | ||
− | Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2})</math> = 3, <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}</math>. | + | Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2})</math> = 3, <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}</math>. - cookiemonster2004 |
===Solution 2=== | ===Solution 2=== |
Revision as of 19:22, 4 April 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, .
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 |