Difference between revisions of "2023 AMC 12A Problems/Problem 10"
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| − | + | = Problem = | |
| + | Positive real numbers <math>x</math> and <math>y</math> satisfy <math>y^3=x^2</math> and <math>(y-x)^2=4y^2</math>. What is <math>x+y</math>? | ||
| + | <math>\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42</math> | ||
| − | - | + | = Solution = |
| + | Because <math>y^3=x^2</math>, set <math>x=a^3</math>, <math>y=a^2</math> (<math>a\neq 0</math>). Put them in <math>(y-x)^2=4y^2</math> we get <math>(a^2(a-1))^2=4a^4</math> which implies <math>a^2-2a+1=4</math>. Solve the equation to get <math>a=3</math> or <math>-1</math>. Since <math>x</math> and <math>y</math> are positive, <math>a=3</math> and <math>x+y=3^3+3^2=\boxed{\textbf{(D)} 36}</math>. | ||
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| + | ~Plasta | ||
Revision as of 19:10, 9 November 2023
Problem
Positive real numbers 
and 
satisfy 
and 
. What is 
?
Solution
Because , set 
, 
(
). Put them in we get 
which implies 
. Solve the equation to get 
or 
. Since and are positive, and 
.
~Plasta
