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 "2023 AMC 12A Problems/Problem 10"

(→‎Solution 2: Quadratic formula)
m (→‎Solution 2: Quadratic formula)
Line 18: Line 18:
  
  
−
== Solution 2: Quadratic formula==  
+
== Solution 3: Quadratic formula==  
 
first expand  
 
first expand  
  

Revision as of 07:56, 10 November 2023

Problem

Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$. What is $x+y$? $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution 1

Because $y^3=x^2$, set $x=a^3$, $y=a^2$ ($a\neq 0$). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$. Solve the equation to get $a=3$ or $-1$. Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{\textbf{(D)} 36}$.

~plasta

Solution 2

Let's take the second equation and square root both sides. This will obtain $y-x = \pm2y$. Solving the case where $y-x=+2y$, we'd find that $x=-y$. This is known to be false because both $x$ and $y$ have to be positive, and $x=-y$ implies that at least one of the variables is not positive. So we instead solve the case where $y-x=-2y$. This means that $x=3y$. Inputting this value into the first equation, we find: \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$. Therefore, $x+y=9+27=\boxed{36}$

~lprado


Solution 3: Quadratic formula

first expand

$(y-x)^2 = 4y^2$

$y^2-2xy+x^2 = 4y^2$

$y^2-2yx+x^2 = 4y^2$

$x^2-2xy-3y^2 = 0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

consider a=1 b=-2y c=-3y^2

$x=\frac{2y\pm\sqrt{(-2y)^2-4(1)(-3y^2)}}{2a}$

$x=\frac{2y\pm\sqrt{16y^2}}{2a}$

$\frac{2y+4y^{2}}{2}$ or $\frac{2y-4y^{2}}{2}$

$x=y+2y$ and $x=y-2y$

$x=3y$ and $x=-y$

we can see both x and y will be postive in $x=3y$

now do same as solution 2 \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$. Therefore, $x+y=9+27=\boxed{36}$

See also

2023 AMC 12A (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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_10&oldid=201792"