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| − | ==Solutions== | + | ==Solution 1== |
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| − | ===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>
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| − | 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 |
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| − | ===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> |
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| − | ===Solution 3===
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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>.
| + | ~idk12345678 |
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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> | + | ==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. |
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| − | 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 |
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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>.
| + | ==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> |
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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>.
| + | ==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> |
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| − | ~kevinmathz | + | ~im_space_cadet |
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| − | ===Solution 4 (Geometric Interpretation)===
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| − | 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>.
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| − | <asy>
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| − | var s = sqrt(3);
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| − | pair A = (-5*s/2, 0);
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| − | pair B = (0,0);
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| − | pair C = (0,5.5);
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| − | pair D = (0,2.5);
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| − | draw(A--B--C--A--D);
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| − | rightanglemark(A, B, D);
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| − | label("A", A, SW);
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| − | label("B", B, SE);
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| − | label("C", C, NE);
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| − | label("D", D, E);
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| − | label("7", (-5*s/4, 5.5/2), NW);
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| − | label("120$^\circ$", D, NW);
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| − | label("60$^\circ$", (0,2), SW);
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| − | label("$x$", 0.5*A, S);
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| − | draw(rightanglemark(A, B, C));
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| − | draw(anglemark(A, D, B));
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| − | markscalefactor = 0.04;
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| − | draw(anglemark(C, D, A));
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| − | label("$\frac{5}{2}$", (0,1.25), E);
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| − | label("3", (0,4), E);
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| − | label("5", (-5*s/4, 5/4), N);
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| − | </asy>
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| − | ===Solution 6 (Symmetric Substitution)===
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| − | Since <math>\frac{25+49}{2}=37</math>, 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{\textbf{(A)}\ 8}</math>.
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| − | ===Solution 7 (Difference of Squares)===
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| − | Let <math>\sqrt{49-x^2}=a</math> and <math>\sqrt{25-x^2}=b</math>. Then by difference of squares:
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| − | <math>(a+b)(a-b)=a^2-b^2</math>.
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| − | 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:
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| − | <math>(a+b)(3)=24</math>.
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| − | Hence, <math>a+b=\sqrt{49-x^2}-\sqrt{25-x^2}=8</math> so our answer is <math>\boxed{\textbf{(A) } 8}</math>.
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| − | ~BakedPotato66
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| − | ===Solution 8 (Analytic Geometry)===
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| − | [[File:2018 AMC10 A P10.PNG|500px]]
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| − | The problem can be represented by the above diagram. The large circle with center <math>O</math> has a radius of 7, the small circle with center <math>O</math> has a radius of 5. Point <math>C</math>'s X coordinate is <math>x</math>. <math>AC=CD=\sqrt{49-x^2}</math>, <math>BC=\sqrt{25-x^2}</math>, <math>AB=AC-BC=\sqrt{49-x^2} - \sqrt{25-x^2} = 3</math>, <math>BD=CD+BC=\sqrt{49-x^2} + \sqrt{25-x^2}</math>.
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| − | By Power of a Point, <math>AB \cdot BD=BE \cdot BF=(7-5) \cdot (7+5)=24</math>, <math>BD=\boxed{\textbf{(A) } 8}</math>
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| − | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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| − | ===Solution 9 (Pythagorean Theorem)===
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| − | Notice that <math>\sqrt{49-x^2} = \sqrt{7^2-x^2}</math> and <math>\sqrt{25-x^2} = \sqrt{5^2-x^2}</math> This is also the equation of finding a leg of a right triangle given the hypotenuse and the other leg using the [[Pythagorean Theorem]].
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| − | Now, <math>7</math> and <math>5</math> are the hypotenuses of the two triangles, and <math>x</math> is the one leg from each of the triangles. So, <math>\sqrt{7^2-x^2}</math> is the other leg of the 1st one, and <math>\sqrt{5^2-x^2}</math> is the other leg of the 2nd one.
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| − | For convenience, we name the other leg of the 1st triangle <math>a</math> (the one that's not <math>5</math> or <math>x</math>), and the other leg of the 2nd one <math>b</math> (the one that's not <math>7</math> or <math>x</math>). Using the Pythagorean Theorem, we set up 2 equations.
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| − | <cmath>\begin{align*}
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| − | a^2 + x^2 &= 7^2 \\
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| − | b^2 + x^2 &= 5^2
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| − | \end{align*}</cmath>
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| − | Subtracting the two equations and canceling out <math>x^2</math>, we have <math>a^2 - b^2 = 49-25</math>, which simplifies to <math>(a-b)(a+b)=24</math>.
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| − | We already know that <math>a-b</math> (or <math>\sqrt{49-x^2}-\sqrt{25-x^2}</math>) is equal to <math>3</math>, so plugging it in, we have <math>3(a+b)=24</math>, and dividing by <math>3</math> gives <math>a+b = \boxed{\textbf{(A)}\ 8}</math>
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| − | ~MrThinker
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| | ==Video Solution (HOW TO THINK CREATIVELY!)== | | ==Video Solution (HOW TO THINK CREATIVELY!)== |
Problem
Suppose that real number 
satisfies ![\[\sqrt{49-x^2}-\sqrt{25-x^2}=3\]](//latex.artofproblemsolving.com/6/0/2/60260471de484d4c7c3af264a757d4ff5de2ad3d.png)
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!)
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
satisfies ![\[\sqrt{49-x^2}-\sqrt{25-x^2}=3\]](http://latex.artofproblemsolving.com/6/0/2/60260471de484d4c7c3af264a757d4ff5de2ad3d.png)
What is the value of 
?
; in other words, we want to find 
. We know that 
Thus, 
.
, and 
. Solving for the constants in terms of 
, 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 
for 
. Moving the 
to the other side and squaring gives us 
, solving for 
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 
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 
