Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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===Solution 6 (Symmetric Substitution)=== | ===Solution 6 (Symmetric Substitution)=== | ||
Since <math>\frac{25+49}{2}=37</math>, let's let <math>37-x^2 = y</math>. Then we have <math>\sqrt{y+12}-\sqrt{y-12}=3</math>. Squaring both sides gives us <math>2y-2\sqrt{y^2-144}=9</math>. Isolating the term with the square root, and squaring again, we get <math>4y^2-36y+81=4y^2-576 \implies y=\frac{73}{4}</math>. Then <math>\sqrt{y+12}+\sqrt{y-12} = \sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}} = \frac{16}{2}=\boxed{8}</math>. | Since <math>\frac{25+49}{2}=37</math>, let's let <math>37-x^2 = y</math>. Then we have <math>\sqrt{y+12}-\sqrt{y-12}=3</math>. Squaring both sides gives us <math>2y-2\sqrt{y^2-144}=9</math>. Isolating the term with the square root, and squaring again, we get <math>4y^2-36y+81=4y^2-576 \implies y=\frac{73}{4}</math>. Then <math>\sqrt{y+12}+\sqrt{y-12} = \sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}} = \frac{16}{2}=\boxed{8}</math>. | ||
− | == Video Solution == | + | == Video Solutions == |
+ | ===Video Solution 1=== | ||
https://youtu.be/ba6w1OhXqOQ?t=1403 | https://youtu.be/ba6w1OhXqOQ?t=1403 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | ===Video Solution 2=== | ||
+ | https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go | ||
+ | |||
+ | ===Video Solution 3=== | ||
+ | https://youtu.be/ZiZVIMmo260 | ||
+ | |||
+ | ===Video Solution 4=== | ||
+ | https://youtu.be/5cA87rbzFdw | ||
+ | |||
+ | ~savannahsolver | ||
== Video Solution == | == Video Solution == |
Revision as of 17:56, 10 May 2021
Contents
[hide]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 . - 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 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: .
~kevinmathz
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 .
-aze.10
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 .
Video Solutions
Video Solution 1
https://youtu.be/ba6w1OhXqOQ?t=1403
~ pi_is_3.14
Video Solution 2
https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go
Video Solution 3
Video Solution 4
~savannahsolver
Video Solution
https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go
Video Solution
~savannahsolver
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 |