Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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− | == | + | ==Problem== |
− | |||
− | + | Suppose that real number <math>x</math> satisfies <cmath>\sqrt{49-x^2}-\sqrt{25-x^2}=3</cmath>What is the value of <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>? | |
− | + | <math> | |
+ | \textbf{(A) }8\qquad | ||
+ | \textbf{(B) }\sqrt{33}+8\qquad | ||
+ | \textbf{(C) }9\qquad | ||
+ | \textbf{(D) }2\sqrt{10}+4\qquad | ||
+ | \textbf{(E) }12\qquad | ||
+ | </math> | ||
− | Solution | + | ==Solution 1== |
− | + | We let <math>a=\sqrt{49-x^2}+\sqrt{25-x^2}</math>; in other words, we want to find <math>a</math>. We know that <math>a\cdot3=\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)\cdot\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=\left(\sqrt{49-x^2}\right)^2-\left(\sqrt{25-x^2}\right)^2=\left(49-x^2\right)-\left(25-x^2\right)=24.</math> Thus, <math>a=\boxed{8}</math>. | |
− | |||
− | + | ~Technodoggo | |
− | + | ==Solution 2== | |
+ | Let <math>a = \sqrt{49-x^2}</math>, and <math>b = \sqrt{25-x^2}</math>. Solving for the constants in terms of x, a , and b, we get <math>a^2 + x^2 = 49</math>, and <math>b^2 + x^2 = 25</math>. Subtracting the second equation from the first gives us <math>a^2 - b^2 = 24</math>. Difference of squares gives us <math>(a+b)(a-b) = 24</math>. Since we want to find <math>a+b = \sqrt{49-x^2}+\sqrt{25-x^2}</math>, and we know <math>a-b = 3</math>, we get <math>3(a+b) = 24</math>, so <math>a+b = \boxed{\textbf{(A) }8}</math> | ||
− | |||
− | + | ~idk12345678 | |
− | + | ==Video Solution (HOW TO THINK CREATIVELY!)== | |
+ | https://youtu.be/P-atxiiTw2I | ||
− | + | ~Education, the Study of Everything | |
+ | |||
+ | |||
+ | |||
+ | |||
+ | == 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=== | ||
+ | https://youtu.be/ZiZVIMmo260 | ||
+ | |||
+ | ===Video Solution 4=== | ||
+ | https://youtu.be/5cA87rbzFdw | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 19:20, 16 May 2024
Contents
Problem
Suppose that real number satisfies What is the value of ?
Solution 1
We let ; in other words, we want to find . We know that Thus, .
~Technodoggo
Solution 2
Let , and . Solving for the constants in terms of x, a , and b, we get , and . Subtracting the second equation from the first gives us . Difference of squares gives us . Since we want to find , and we know , we get , so
~idk12345678
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
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
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 |