Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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− | Suppose that real number <math>x</math> satisfies <cmath>\sqrt{49-x^2}-\sqrt{25-x^2}=3</cmath> | + | ==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> | <math> | ||
− | \textbf{(A) }8 \qquad | + | \textbf{(A) }8\qquad |
\textbf{(B) }\sqrt{33}+8\qquad | \textbf{(B) }\sqrt{33}+8\qquad | ||
− | \textbf{(C) }9 \qquad | + | \textbf{(C) }9\qquad |
− | \textbf{(D) }2\sqrt{10}+4 \qquad | + | \textbf{(D) }2\sqrt{10}+4\qquad |
− | \textbf{(E) }12 \qquad | + | \textbf{(E) }12\qquad |
</math> | </math> | ||
− | == Solution 1== | + | ==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 <math>x</math>, <math>a</math> , and <math>b</math>, 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 | ||
− | + | ==Solution 3== | |
+ | We can substitute <math>25 - x^2</math> for <math>a</math>, thus turning the equation into <math>\sqrt{a+24} - \sqrt{a} = 3</math>. Moving the <math>\sqrt{a}</math> to the other side and squaring gives us <math>a + 24 = 9 + 6\sqrt{a} + a</math>, solving for <math>a</math> gives us 25/4. We substitute this value into the expression they asked us to evaluate giving 8. | ||
− | + | ~ SAMANTAP | |
− | + | ==Solution 4== | |
− | ==Solution | + | Move <math>-\sqrt{25-x^2}</math> to the right to get <math>\sqrt{49-x^2} = 3 + \sqrt{25-x^2}</math>. |
− | + | Square both sides to get <math>49-x^2 = 9 + 6\sqrt{25-x^2} + (25-x^2)</math>. | |
+ | Simplify to get <math>15 = 6\sqrt{25-x^2}</math>, or <math>\frac{5}{2} = \sqrt{25-x^2}</math> | ||
+ | Substitute this back into the original equation tog et that <math>\sqrt{49-x^2} = \frac{11}{2}</math>. The answer is <math>\boxed{\textbf{(A) }8}</math> | ||
− | == See Also == | + | ==Solution 5(Jaideep's Difference of Roots Equals Integer Method)[JDRIM]== |
+ | We are given that, <math>\sqrt(49-x^2) - \sqrt(25-x^2) = 3</math> | ||
+ | We are asked to find, <math>\sqrt(49-x^2) + \sqrt(25-x^2)</math> | ||
+ | Notice that these two expressions are conjugates of one another. Therefore, we can find that by multiply these two conjugates by one another we should be able to find that: <math>(\sqrt(49-x^2) - \sqrt(25-x^2))(\sqrt(49-x^2) + \sqrt(25-x^2)) = (49-x^2) - (25-x^2) </math> | ||
+ | <math>\Rightarrow 49-x^2-25+x^2 = 24</math> | ||
+ | We are already given that the first expression equals 3, thus, our expression now becomes: <math>3(\sqrt(49-x^2)+\sqrt(25-x^2)) = 24 </math> <math>\Rightarrow \sqrt(49-x^2)+\sqrt(25-x^2) = 8 </math> | ||
+ | Thus, the answer is <math>\boxed{\textbf{(A) }8}</math> | ||
+ | |||
+ | ~im_space_cadet | ||
+ | |||
+ | ==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}} | {{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}} | ||
− | + | ||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 17:46, 2 November 2024
Contents
[hide]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 , , and , 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
Solution 3
We can substitute for , thus turning the equation into . Moving the to the other side and squaring gives us , solving for gives us 25/4. We substitute this value into the expression they asked us to evaluate giving 8.
~ SAMANTAP
Solution 4
Move to the right to get . Square both sides to get . Simplify to get , or Substitute this back into the original equation tog et that . The answer is
Solution 5(Jaideep's Difference of Roots Equals Integer Method)[JDRIM]
We are given that, We are asked to find, Notice that these two expressions are conjugates of one another. Therefore, we can find that by multiply these two conjugates by one another we should be able to find that: We are already given that the first expression equals 3, thus, our expression now becomes: Thus, the answer is
~im_space_cadet
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 |