Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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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> | 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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Revision as of 22:52, 9 February 2018
Solutions
Solution 1
In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The 
terms cancel nicely. 
Given that 
= 3, 
Solution by PancakeMonster2004, explanations added by a1b2.
Solution 2 (bad)
Let 
, and let 
. Then 
. Substituting, we get 
. Rearranging, we get 
. Squaring both sides and solving, we get 
and 
. Adding, we get that the answer is 
Solution 3
Put the equations to one side. 
can be changed into 
.
We can square both sides, getting us 
That simplifies out to 
Dividing both sides gets us 
.
Following that, we can square both sides again, resulting in the equation 
. Simplifying that, we get 
.
Substituting into the equation 
, we get 
. Immediately, we simplify into 
. The two numbers inside the square roots are simplified to be 
and 
, so you add them up: 
~kevinmathz
