Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 5 (No Square Roots, Fastest))
m (→‎Solution 2)
 
(32 intermediate revisions by 12 users not shown)
Line 11: Line 11:
 
</math>
 
</math>
  
−
==Solutions==
+
==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>.
−
In order to eliminate 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>
 
  
−
Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3, (\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}</math>. - cookiemonster2004
+
~Technodoggo
  
−
===Solution 2===
+
==Solution 2==
−
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>.
+
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>
  
−
===Solution 3===
 
  
−
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>.
+
~idk12345678
  
−
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>
+
==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.
  
−
That simplifies out to <math>15=6 \sqrt{25-x^2}.</math> Dividing both sides by <math>6</math>  gets us <math>\frac{5}{2}=\sqrt{25-x^2}</math>.
+
~ SAMANTAP
  
−
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>.
+
==Solution 4==
 +
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>
  
−
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>.
+
==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>
  
−
~kevinmathz
+
~im_space_cadet
  
−
===Solution 4 (Geometric Interpretation)===
+
==Video Solution (HOW TO THINK CREATIVELY!)==
 +
https://youtu.be/P-atxiiTw2I
  
−
Draw a right triangle <math>ABC</math> with a hypotenuse <math>AC</math> of length <math>7</math> and leg <math>AB</math> of length <math>x</math>. Draw <math>D</math> on <math>BC</math> such that <math>AD=5</math>. Note that <math>BC=\sqrt{49-x^2}</math> and <math>BD=\sqrt{25-x^2}</math>. Thus, from the given equation, <math>BC-BD=DC=3</math>. Using Law of Cosines on triangle <math>ADC</math>, we see that <math>\angle{ADC}=120^{\circ}</math> so <math>\angle{ADB}=60^{\circ}</math>. Since <math>ADB</math> is a <math>30-60-90</math> triangle, <math>\sqrt{25-x^2}=BD=\frac{5}{2}</math> and <math>\sqrt{49-x^2}=\frac{5}{2}+3=\frac{11}{2}</math>. Finally, <math>\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{5}{2}+\frac{11}{2}=\boxed{\textbf{(A)~8}}</math>.
+
~Education, the Study of Everything
−
<asy>
 
−
var s = sqrt(3);
 
−
pair A = (-5*s/2, 0);
 
−
pair B = (0,0);
 
−
pair C = (0,5.5);
 
−
pair D = (0,2.5);
 
  
−
draw(A--B--C--A--D);
 
−
rightanglemark(A, B, D);
 
−
label("A", A, SW);
 
−
label("B", B, SE);
 
−
label("C", C, NE);
 
−
label("D", D, E);
 
−
label("7", (-5*s/4, 5.5/2), NW);
 
−
label("120$^\circ$", D, NW);
 
−
label("60$^\circ$", (0,2), SW);
 
−
label("$x$", 0.5*A, S);
 
−
draw(rightanglemark(A, B, C));
 
  
−
draw(anglemark(A, D, B));
 
−
markscalefactor = 0.04;
 
−
draw(anglemark(C, D, A));
 
  
−
label("$\frac{5}{2}$", (0,1.25), E);
 
−
label("3", (0,4), E);
 
−
label("5", (-5*s/4, 5/4), N);
 
−
</asy>
 
−
 
−
===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 it as <math>((49-x^2) - (25 - x^2)) = (49 - x^2 - 25 + x^2) = 24</math>. Given the <math>3</math> in the problem, we can divide <math>24 / 3 = \boxed{\textbf{(A) } 8}</math>.
 
−
 
−
-aze.10
 
−
 
−
===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>.
 
−
 
−
===Solution 7 (Difference of Squares)===
 
−
Let <math>\sqrt{49-x^2}=a</math> and <math>\sqrt{25-x^2}=b</math>. Then by difference of squares:
 
−
 
−
<math>(a+b)(a-b)=a^2-b^2</math>.
 
−
 
−
We can simplify this expression to get our answer. <math>a^2-b^2=(49-x^2)-(25-x^2)=24</math> and from the given statement, <math>a-b=3</math>. Now we have:
 
−
 
−
<math>(a+b)(3)=24</math>.
 
−
 
−
Hence, <math>a+b=\sqrt{49-x^2}-\sqrt{25-x^2}=8</math> so our answer is <math>\boxed{\textbf{(A) } 8}</math>.
 
−
 
−
~BakedPotato66
 
  
 
== Video Solutions ==
 
== Video Solutions ==

Latest revision as of 17:46, 2 November 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_10&oldid=231267"