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2018 AMC 10A Problems/Problem 10: Difference between revisions

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~Technodoggo
~Technodoggo
Alternative method:
Let <math>a=\sqrt{49-x^2}+\sqrt{25-x^2}</math>, <math>b=\sqrt{49-x^2}-\sqrt{25-x^2}</math>. <math>a*b=49-x^2 - 25 + x^2 = 24</math>. The problem states that <math>b=3</math>, so we get <math>a = 24 / 3 = 8 </math>. The answer is <math>\boxed{\textbf{(A) }8}</math>.
~SwordAxe


==Solution 2==
==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>
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>




Line 34: Line 39:
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>
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>


==Solution 5(Jaideep's Difference of Roots Equals Integer Method)[JDRIM]
==Solution 5(Jaideep's Difference of Roots Equals Integer Method)[JDRIM]==
We are given that,  
We are given that, <math>\sqrt(49-x^2) - \sqrt(25-x^2) = 3</math>
                        <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>
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: <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>\sqrt(49-x^2) + \sqrt(25-x^2)</math>
<math>\Rightarrow 49-x^2-25+x^2 = 24</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:
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>
                        Difference of Squares Formula: <math>(a+b)(a-b)=a^2-b^2</math>
                        <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>
Thus, the answer is <math>\boxed{\textbf{(A) }8}</math>


~im_space_cadet
~im_space_cadet
 
 
~Failure.net


==Video Solution (HOW TO THINK CREATIVELY!)==
==Video Solution (HOW TO THINK CREATIVELY!)==

Latest revision as of 20:14, 3 November 2025

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

Alternative method: Let $a=\sqrt{49-x^2}+\sqrt{25-x^2}$, $b=\sqrt{49-x^2}-\sqrt{25-x^2}$. $a*b=49-x^2 - 25 + x^2 = 24$. The problem states that $b=3$, so we get $a = 24 / 3 = 8$. The answer is $\boxed{\textbf{(A) }8}$.

~SwordAxe

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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