Difference between revisions of "2023 AMC 12A Problems/Problem 10"

(Solution 2: Quadractic formula)
 
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== Solution 2: Quadractic formula==  
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== Solution 3: Quadratic formula==  
 
first expand  
 
first expand  
  
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<math>x=\frac{2y\pm\sqrt{16y^2}}{2a}</math>
 
<math>x=\frac{2y\pm\sqrt{16y^2}}{2a}</math>
  
<math>x=\frac{2y\pm\4y}}{2a}</math>
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<math>\frac{2y+4y^{2}}{2}</math> or <math>\frac{2y-4y^{2}}{2}</math>
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<math>x=y+2y</math> and <math>x=y-2y</math>
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<math>x=3y</math> and <math>x=-y</math>
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we can see both x and y will be postive in <math>x=3y</math>
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now do same as solution 2
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<cmath>y^3 = (3y)^2</cmath>
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<cmath>y^3 = 9y^2</cmath>
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<cmath>y=9</cmath>
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This means that <math>x=3y=3(9)=27</math>. Therefore, <math>x+y=9+27=\boxed{36}</math>
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== Solution 4: Substitution==
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Since <math>a^2 = |a|^2</math>, we can rewrite the second equation as <math>(x-y)^2=4y^2</math>
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Let <math>u=x+y</math>. The second equation becomes
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<cmath>(u-2y)^2 = 4y^2</cmath>
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<cmath>u^2 - 4uy = 0</cmath>
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<cmath>u = 4y</cmath>
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<cmath>x+y = 4y</cmath>
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<cmath>x = 3y.</cmath>
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Substituting this into the first equation, we have
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<cmath>y^3 = (3y)^2,</cmath>
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so <math>x = 9</math>.
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Hence <math>x = 27</math> and <math>x + y = \boxed{\textbf{(D)} 36}.</math>
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-Benedict T (countmath1)
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== Solution 5: Difference of Squares==
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We will use the difference of squares in the second equation.
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<cmath>(y-x)^2=4y^2</cmath>
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<cmath>(y-x)^2-(2y)^2=0</cmath>
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<cmath>(y-x-2y)(y-x+2y)=0</cmath>
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<cmath>-(x+y)(3y-x)=0</cmath>
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 +
Since x and y are positive, x+y is non-zero. Thus, <cmath>3y=x</cmath>.
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 +
Substituting into the first equation:
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<cmath>y^3=x^2</cmath>
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<cmath>y^3=9y^2</cmath>
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<cmath>y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}</cmath>
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 +
==Video Solution (easy to digest) by Power Solve==
 +
https://youtu.be/YXIH3UbLqK8?si=ia236Rh1UU681RNT&t=1279
 +
 
 +
==Video Solution (⚡ Under 3 minutes ⚡)==
 +
https://youtu.be/G63hRr9bkzI
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/OqlerYo-uPU
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See also==
 
==See also==

Latest revision as of 19:35, 10 June 2024

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, $x+y=9+27=\boxed{36}$

Solution 4: Substitution

Since $a^2 = |a|^2$, we can rewrite the second equation as $(x-y)^2=4y^2$

Let $u=x+y$. The second equation becomes


\[(u-2y)^2 = 4y^2\] \[u^2 - 4uy = 0\] \[u = 4y\] \[x+y = 4y\] \[x = 3y.\]

Substituting this into the first equation, we have

\[y^3 = (3y)^2,\] so $x = 9$.

Hence $x = 27$ and $x + y = \boxed{\textbf{(D)} 36}.$

-Benedict T (countmath1)

Solution 5: Difference of Squares

We will use the difference of squares in the second equation.

\[(y-x)^2=4y^2\] \[(y-x)^2-(2y)^2=0\] \[(y-x-2y)(y-x+2y)=0\] \[-(x+y)(3y-x)=0\]

Since x and y are positive, x+y is non-zero. Thus, \[3y=x\].

Substituting into the first equation:

\[y^3=x^2\] \[y^3=9y^2\] \[y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}\]

Video Solution (easy to digest) by Power Solve

https://youtu.be/YXIH3UbLqK8?si=ia236Rh1UU681RNT&t=1279

Video Solution (⚡ Under 3 minutes ⚡)

https://youtu.be/G63hRr9bkzI

~Education, the Study of Everything

Video Solution

https://youtu.be/OqlerYo-uPU

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AMC 12A (ProblemsAnswer KeyResources)
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

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