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Difference between revisions of "2024 AMC 12A Problems/Problem 15"

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We find that <math>f(2i)=-8i-8-2i+3=-10i-5</math> and <math>f(-2i)=8i-8+2i+3=10i-5</math>, so <math>f(2i)f(-2i)=(-5-10i)(-5+10i)=(-5)^2-(10i)^2=25+100=\boxed{\textbf{(D) }125}</math>.
 
We find that <math>f(2i)=-8i-8-2i+3=-10i-5</math> and <math>f(-2i)=8i-8+2i+3=10i-5</math>, so <math>f(2i)f(-2i)=(-5-10i)(-5+10i)=(-5)^2-(10i)^2=25+100=\boxed{\textbf{(D) }125}</math>.
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 +
Select D
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 +
  
 
~eevee9406
 
~eevee9406
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==Solution 4 (Reduction of power)==
 
==Solution 4 (Reduction of power)==
  
−
The motivation for this solution is the observation that <math>(p+c)(q+c)(r+c)</math> is easy to compute for any constant c, since <math>(p+c)(q+c)(r+c)=-f(-c)</math>, where <math>f</math> is the polynomial given in the problem. The idea is to transform the expression involving <math>p^2, q^2, r^2</math> into one involving <math>p, q, r</math>.
+
The motivation for this solution is the observation that <math>(p+c)(q+c)(r+c)</math> is easy to compute for any constant c, since <math>(p+c)(q+c)(r+c)=-f(-c)</math> (*), where <math>f</math> is the polynomial given in the problem. The idea is to transform the expression involving <math>p^2, q^2, r^2</math> into one involving <math>p, q, r</math>.
 +
 
 +
Since <math>p</math> is a root of <math>f</math>,
 +
<cmath>p^3+2p^2-p+3=0\implies p^3+2p^2=p^2(p+2)=p-3,</cmath>
 +
which gives us that <math>p^2=\frac{p-3}{p+2}</math>. Then
 +
<cmath>p^2+4=\frac{p-3}{p+2} + 4 = \frac{5p+5}{p+2}=5\cdot \frac{p+1}{p+2}.</cmath>
 +
Since <math>q</math> and <math>r</math> are also roots of <math>f</math>, the same analysis holds, so
 +
 
 +
\begin{align*}
 +
(p^2 + 4)(q^2 + 4)(r^2 + 4)&= \left(5\cdot \frac{p+1}{p+2}\right)\left(5\cdot \frac{q+1}{q+2}\right)\left(5\cdot \frac{r+1}{r+2}\right)\\
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&= 125 \frac{(p+1)(q+1)(r+1)}{(p+2)(q+2)(r+2)}\\
 +
&= 125 \frac{-f(-1)}{-f(-2)} \\
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&= 125\cdot 1\\
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&=\boxed{\textbf{(D) }125}.
 +
\end{align*}
 +
 
 +
(*) This is because
 +
<cmath>(p+c)(r+c)(q+c)=(-1)^3(-c-p)(-c-r)(-c-q)=-f(-c),</cmath>
 +
since
 +
<cmath>f(x)=(x-p)(x-q)(x-r)</cmath>
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for all <math>x</math>.
  
−
Since <math>p^3+2px^2-p+3=0</math>, we get <math>p^2=\frac{p-3}{p+2}</math>. Similarly <math>q^2=\frac{q-3}{q+2}</math> and <math>r^2=\frac{r-3}{r+2}</math>.
+
~tsun26
 +
~KSH31415 (final step and clarification)
  
−
Hence,
+
==Solution 5 (Cheesing it out)==
 +
Expanding the expression
 
<cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4)</cmath>
 
<cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4)</cmath>
−
<cmath>=\left(\frac{p-3}{p+2}+4\right)\left(\frac{q-3}{q+2}+4\right)\left(\frac{r-3}{r+2}+4\right)</cmath>
+
gives us
−
<cmath>=\left(\frac{5p+5}{p+1}\right)\left(\frac{5q+5}{q+1}\right)\left(\frac{5r+5}{r+1}\right)</cmath>
+
<cmath>(pqr)^2+4p^2q^2+4p^2r^2+4q^2r^2+16p^2+16q^2+16r^2+64</cmath>
−
<cmath>=125</cmath>
+
 
 +
Notice that everything other than <math>(pqr)^2</math> is a multiple of <math>4</math>. Solving for <math>(pqr)^2</math> using vieta's formulas, we get <math>9</math>. Since <math>9</math> is <math>1\pmod4</math>, the answer should be as well. The only answer that is <math>1\pmod4</math> is <math>\boxed{\textbf{(D) }125}</math>.
 +
 
 +
~callyaops
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 +
 
 +
==Solution 6==
 +
 
 +
Suppose <math>y = x^2 + 4</math>
 +
 
 +
then <math>x =\pm \sqrt{y - 4}</math>. Substitute <math>x = \sqrt{y - 4}</math> into <math>x^3 + 2x^2 - x + 3 = 0</math> (It is same for <math>x = -\sqrt{y - 4}</math> because the squares in <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)</math>)
 +
 
 +
<math>(\sqrt{y - 4})^3 + 2(\sqrt{y - 4})^2 - \sqrt{y - 4} + 3 = 0 \implies (y - 5)^2(y - 4) = (-2y + 5)^2</math> whose constant is 125
 +
 
 +
according to Vieta's theorem, <math>y_1y_2y_3 = 125</math>
  
−
Interestingly, we didn't even need to calculate the numerator and denominator as they just cancel out. Our answer is <math>\boxed{\textbf{(D) }125}</math>.  
+
<math>y_1y_2y_3 = 125 \implies (x_1^2 + 4)(x_2^2 + 4)(x_3^2 + 4) = 125 \implies (p^2 + 4)(q^2 + 4)(r^2 + 4) = \boxed{\textbf{(D) }125}</math>.  
  
−
~tsun26
+
~JiYang
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}}
 
{{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:03, 18 November 2024

Problem

The roots of $x^3 + 2x^2 - x + 3$ are $p, q,$ and $r.$ What is the value of \[(p^2 + 4)(q^2 + 4)(r^2 + 4)?\]$\textbf{(A) } 64 \qquad \textbf{(B) } 75 \qquad \textbf{(C) } 100 \qquad \textbf{(D) } 125 \qquad \textbf{(E) } 144$

Solution 1

You can factor $(p^2 + 4)(q^2 + 4)(r^2 + 4)$ as $(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)$.

For any polynomial $f(x)$, you can create a new polynomial $f(x+2)$, which will have roots that instead have the value subtracted.

Substituting $x-2$ and $x+2$ into $x$ for the first polynomial, gives you $10i-5$ and $-10i-5$ as $c$ for both equations. Multiplying $10i-5$ and $-10i-5$ together gives you $\boxed{\textbf{(D) }125}$.

-ev2028

~Latex by eevee9406

Solution 2

Let $f(x)=x^3 + 2x^2 - x + 3$. Then $(p^2 + 4)(q^2 + 4)(r^2 + 4)=(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)=f(2i)f(-2i)$.


We find that $f(2i)=-8i-8-2i+3=-10i-5$ and $f(-2i)=8i-8+2i+3=10i-5$, so $f(2i)f(-2i)=(-5-10i)(-5+10i)=(-5)^2-(10i)^2=25+100=\boxed{\textbf{(D) }125}$.

Select D


~eevee9406

Solution 3

First, denote that \[p+q+r=-2, pq+pr+qr=-1, pqr=-3\] Then we expand the expression \[(p^2+4)(q^2+4)(r^2+4)\] \[=(pqr)^2+4((pq)^2+(pr)^2+(qr)^2)+4^2(p^2+q^2+r^2)+4^3\] \[=(-3)^2+4((pq+pr+qr)^2-2pqr(p+q+r))+4^2((p+q+r)^2-2(pq+pr+qr))+4^3\] \[=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3\] \[=\fbox{(D) 125}\] ~lptoggled

Solution 4 (Reduction of power)

The motivation for this solution is the observation that $(p+c)(q+c)(r+c)$ is easy to compute for any constant c, since $(p+c)(q+c)(r+c)=-f(-c)$ (*), where $f$ is the polynomial given in the problem. The idea is to transform the expression involving $p^2, q^2, r^2$ into one involving $p, q, r$.

Since $p$ is a root of $f$, \[p^3+2p^2-p+3=0\implies p^3+2p^2=p^2(p+2)=p-3,\] which gives us that $p^2=\frac{p-3}{p+2}$. Then \[p^2+4=\frac{p-3}{p+2} + 4 = \frac{5p+5}{p+2}=5\cdot \frac{p+1}{p+2}.\] Since $q$ and $r$ are also roots of $f$, the same analysis holds, so

\begin{align*} (p^2 + 4)(q^2 + 4)(r^2 + 4)&= \left(5\cdot \frac{p+1}{p+2}\right)\left(5\cdot \frac{q+1}{q+2}\right)\left(5\cdot \frac{r+1}{r+2}\right)\\ &= 125 \frac{(p+1)(q+1)(r+1)}{(p+2)(q+2)(r+2)}\\ &= 125 \frac{-f(-1)}{-f(-2)} \\ &= 125\cdot 1\\ &=\boxed{\textbf{(D) }125}. \end{align*}

(*) This is because \[(p+c)(r+c)(q+c)=(-1)^3(-c-p)(-c-r)(-c-q)=-f(-c),\] since \[f(x)=(x-p)(x-q)(x-r)\] for all $x$.

~tsun26 ~KSH31415 (final step and clarification)

Solution 5 (Cheesing it out)

Expanding the expression \[(p^2 + 4)(q^2 + 4)(r^2 + 4)\] gives us \[(pqr)^2+4p^2q^2+4p^2r^2+4q^2r^2+16p^2+16q^2+16r^2+64\]

Notice that everything other than $(pqr)^2$ is a multiple of $4$. Solving for $(pqr)^2$ using vieta's formulas, we get $9$. Since $9$ is $1\pmod4$, the answer should be as well. The only answer that is $1\pmod4$ is $\boxed{\textbf{(D) }125}$.

~callyaops


Solution 6

Suppose $y = x^2 + 4$

then $x =\pm \sqrt{y - 4}$. Substitute $x = \sqrt{y - 4}$ into $x^3 + 2x^2 - x + 3 = 0$ (It is same for $x = -\sqrt{y - 4}$ because the squares in $(p^2 + 4)(q^2 + 4)(r^2 + 4)$)

$(\sqrt{y - 4})^3 + 2(\sqrt{y - 4})^2 - \sqrt{y - 4} + 3 = 0 \implies (y - 5)^2(y - 4) = (-2y + 5)^2$ whose constant is 125

according to Vieta's theorem, $y_1y_2y_3 = 125$

$y_1y_2y_3 = 125 \implies (x_1^2 + 4)(x_2^2 + 4)(x_3^2 + 4) = 125 \implies (p^2 + 4)(q^2 + 4)(r^2 + 4) = \boxed{\textbf{(D) }125}$.

~JiYang

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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