2018 AMC 10A Problems/Problem 10
- 1 Problem
- 2 Solutions
- 3 Video Solutions
- 4 See Also
Suppose that real number satisfies What is the value of ?
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 . - cookiemonster2004
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 .
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 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, .
Solution 5 (No Square Roots, Fastest)
We notice that the two expressions are conjugates, and therefore we can write them in a "difference-of-squares" format. Namely, we can write is as . Given the in the problem, we can divide .
Solution 6 (Symmetric Substitution)
Since , let's let . Then we have . Squaring both sides gives us . Isolating the term with the square root, and squaring again, we get . Then .
Solution 7 (Difference of Squares)
Let and . Then by difference of squares:
We can simplify this expression to get our answer. and from the given statement, . Now we have:
Hence, so our answer is .
Video Solution 1
Video Solution 2
https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go
Video Solution 3
Video Solution 4
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