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Difference between revisions of "2018 AMC 10A Problems/Problem 10"

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==Solutions==
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==Problem==
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=== Solution 1===
 
  
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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>
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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>?
  
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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>
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<math>
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\textbf{(A) }8\qquad
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\textbf{(B) }\sqrt{33}+8\qquad
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\textbf{(C) }9\qquad
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\textbf{(D) }2\sqrt{10}+4\qquad
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\textbf{(E) }12\qquad
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</math>
  
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Solution by PancakeMonster2004, explanations added by a1b2.
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==Solution 1==
  
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===Solution 2 ===
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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>.
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Let <math>u=\sqrt{49-x^2}</math>, and let <math>v=\sqrt{25-x^2}</math>. Then <math>v=\sqrt{u^2-24}</math>. Substituting, we get <math>u-\sqrt{u^2-24}=3</math>. Rearranging, we get <math>u-3=\sqrt{u^2-24}</math>. Squaring both sides and solving, we get <math>u=\frac{11}{2}</math> and <math>v=\frac{11}{2}-3=\frac{5}{2}</math>. Adding, we get that the answer is <math>\boxed{\textbf{(A) } 8}</math>
 
  
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===Solution 3===
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~Technodoggo
  
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Put the equations to one side. <math>\sqrt{49-x^2}-\sqrt{25-x^2}=3</math> can be changed into <math>\sqrt{49-x^2}=\sqrt{25-x^2}+3</math>.
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==Solution 2==
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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>
  
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We can square both sides, getting us <math>49-x^2=(25-x^2)+(3^2)+ 2\cdot 3 \cdot \sqrt{25-x^2}.</math>
 
  
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That simplifies out to <math>15=6 \sqrt{25-x^2}.</math> Dividing both sides gets us <math>\frac{5}{2}=\sqrt{25-x^2}</math>.
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~idk12345678
  
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Following that, we can square both sides again, resulting in the equation <math>\frac{25}{4}=25-x^2</math>. Simplifying that, we get <math>x^2 = \frac{75}{4}</math>.
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==Solution 3==
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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.
  
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Substituting into the equation <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>, we get <math>\sqrt{49-\frac{75}{4}}+\sqrt{25-\frac{75}{4}}</math>. Immediately, we simplify into <math>\sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}}</math>. The two numbers inside the square roots are simplified to be <math>\frac{11}{2}</math> and <math>\frac{5}{2}</math>, so you add them up: <math>\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}</math>
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~ SAMANTAP
  
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~kevinmathz
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==Solution 4==
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Move <math>-\sqrt{25-x^2}</math> to the right to get <math>\sqrt{49-x^2} = 3 + \sqrt{25-x^2}</math>.
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Square both sides to get <math>49-x^2 = 9 + 6\sqrt{25-x^2} + (25-x^2)</math>.
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Simplify to get <math>15 = 6\sqrt{25-x^2}</math>, or <math>\frac{5}{2} = \sqrt{25-x^2}</math>
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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>
  
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{{AMC10 box|year=2018|ab=A|num-b=9|after=Problem 11}}
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==Solution 5(Jaideep's Difference of Roots Equals Integer Method)[JDRIM]==
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We are given that, <math>\sqrt(49-x^2) - \sqrt(25-x^2) = 3</math>
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We are asked to find, <math>\sqrt(49-x^2) + \sqrt(25-x^2)</math>
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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>
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<math>\Rightarrow 49-x^2-25+x^2 = 24</math>
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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>
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Thus, the answer is <math>\boxed{\textbf{(A) }8}</math>
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~im_space_cadet
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==Video Solution (HOW TO THINK CREATIVELY!)==
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https://youtu.be/P-atxiiTw2I
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
 
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== Video Solutions ==
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===Video Solution 1===
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https://youtu.be/ba6w1OhXqOQ?t=1403
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~ pi_is_3.14
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===Video Solution 2===
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https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go
 +
 
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===Video Solution 3===
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https://youtu.be/ZiZVIMmo260
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===Video Solution 4===
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https://youtu.be/5cA87rbzFdw
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 +
~savannahsolver
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==See Also==
 +
 
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{{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}}
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[[Category:Introductory Algebra Problems]]

Revision as of 22:33, 24 August 2024

Problem

Suppose that real number $x$ satisfies \[\sqrt{49-x^2}-\sqrt{25-x^2}=3\]What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?

$\textbf{(A) }8\qquad \textbf{(B) }\sqrt{33}+8\qquad \textbf{(C) }9\qquad \textbf{(D) }2\sqrt{10}+4\qquad \textbf{(E) }12\qquad$

Solution 1

We let $a=\sqrt{49-x^2}+\sqrt{25-x^2}$; in other words, we want to find $a$. We know that $a\cdot3=\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)\cdot\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=\left(\sqrt{49-x^2}\right)^2-\left(\sqrt{25-x^2}\right)^2=\left(49-x^2\right)-\left(25-x^2\right)=24.$ Thus, $a=\boxed{8}$.

~Technodoggo

Solution 2

Let $a = \sqrt{49-x^2}$, and $b = \sqrt{25-x^2}$. Solving for the constants in terms of x, a , and b, we get $a^2 + x^2 = 49$, and $b^2 + x^2 = 25$. Subtracting the second equation from the first gives us $a^2 - b^2 = 24$. Difference of squares gives us $(a+b)(a-b) = 24$. Since we want to find $a+b = \sqrt{49-x^2}+\sqrt{25-x^2}$, and we know $a-b = 3$, we get $3(a+b) = 24$, so $a+b = \boxed{\textbf{(A) }8}$


~idk12345678

Solution 3

We can substitute $25 - x^2$ for $a$, thus turning the equation into $\sqrt{a+24} - \sqrt{a} = 3$. Moving the $\sqrt{a}$ to the other side and squaring gives us $a + 24 = 9 + 6\sqrt{a} + a$, solving for $a$ gives us 25/4. We substitute this value into the expression they asked us to evaluate giving 8.

~ SAMANTAP

Solution 4

Move $-\sqrt{25-x^2}$ to the right to get $\sqrt{49-x^2} = 3 + \sqrt{25-x^2}$. Square both sides to get $49-x^2 = 9 + 6\sqrt{25-x^2} + (25-x^2)$. Simplify to get $15 = 6\sqrt{25-x^2}$, or $\frac{5}{2} = \sqrt{25-x^2}$ Substitute this back into the original equation tog et that $\sqrt{49-x^2} = \frac{11}{2}$. The answer is $\boxed{\textbf{(A) }8}$

Solution 5(Jaideep's Difference of Roots Equals Integer Method)[JDRIM]

We are given that, $\sqrt(49-x^2) - \sqrt(25-x^2) = 3$ We are asked to find, $\sqrt(49-x^2) + \sqrt(25-x^2)$ 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: $(\sqrt(49-x^2) - \sqrt(25-x^2))(\sqrt(49-x^2) + \sqrt(25-x^2)) = (49-x^2) - (25-x^2)$ $\Rightarrow 49-x^2-25+x^2 = 24$ We are already given that the first expression equals 3, thus, our expression now becomes: $3(\sqrt(49-x^2)+\sqrt(25-x^2)) = 24$ $\Rightarrow \sqrt(49-x^2)+\sqrt(25-x^2) = 8$ Thus, the answer is $\boxed{\textbf{(A) }8}$

~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

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
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