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

m (Reverted edits by Maa is stupid (talk) to last revision by Jiyang)
(Tag: Rollback)
(Solution 2: Clarify function in terms of roots)
 
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==Solution 2==
 
==Solution 2==
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Let <math>f(x)=x^3 + 2x^2 - x + 3</math>. Then
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Let <math>f(x) = x^3 + 2x^2 - x + 3 = (x - p)(x - q)(x - r) = -(p - x)(q - x)(r - x)</math>. Then
 
<math>(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)</math>.
 
<math>(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)</math>.
  
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~lptoggled
 
~lptoggled
  
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==Solution 4 (Reduction of power)==
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==Solution 4 (Newton's Sums and Vieta's)==
 +
Expand the expression:
 +
 
 +
<cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4) = (p^2q^2r^2) + 4(q^2r^2 + p^2r^2 + p^2q^2) + 16(p^2 + q^2 + r^2) + 64.</cmath>
 +
 
 +
By Vieta's we have <math>pqr = -3.</math> Therefore, <math>p^2q^2r^2 = (-3)^2 = 9</math>. We use Newton's Sums to quickly compute <math>p^2 + q^2 + r^2:</math>
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 +
<cmath>S_2 + 2S_1 + 2(-1) = 0</cmath>
 +
 
 +
By Vieta's again, <math>S_1 = p + q + r -2</math> which means <math>S_2 = p^2 + q^2 + r^2 = 6.</math>
 +
 
 +
Now all we need is <math>q^2r^2 + p^2r^2 + p^2q^2.</math> Vieta's also tells us that <math>pq + qr + pr = -1,</math> so we can take this equation and square both sides:
 +
 
 +
<cmath>(pq + qr + pr)^2 = q^2r^2 + p^2r^2 + p^2q^2 + 2(qpr^2 + q^2rp + p^2rq) = 1.</cmath>
 +
 
 +
We know <math>pqr = -3</math> so we can substitute this in:
 +
 
 +
<cmath>(pq + qr + pr)^2 = q^2r^2 + p^2r^2 + p^2q^2 -6(r + p + q) = 1.</cmath>
 +
 
 +
We also know <math>p + q + r = -2</math>, which means <math>q^2r^2 + p^2r^2 + p^2q^2 = -11.</math>
 +
 
 +
 
 +
So the answer is just <math>9 + 4(-11) + 16(6) + 64 = \boxed{125}.</math>
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~[[User:grogg007|grogg007]]
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 +
==Solution 5 (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>.
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~KSH31415 (final step and clarification)
 
~KSH31415 (final step and clarification)
  
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==Solution 5 (Cheesing it out)==
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==Solution 6 (Cheesing it out)==
 
Expanding the expression
 
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>
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~callyaops
 
~callyaops
  
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==Solution 7==
−
==Solution 6==
 
  
 
Suppose <math>y = x^2 + 4</math>
 
Suppose <math>y = x^2 + 4</math>

Latest revision as of 00:30, 13 August 2025

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 $f(x) = x^3 + 2x^2 - x + 3 = (x - p)(x - q)(x - r) = -(p - x)(q - x)(r - x)$. 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 (Newton's Sums and Vieta's)

Expand the expression:

\[(p^2 + 4)(q^2 + 4)(r^2 + 4) = (p^2q^2r^2) + 4(q^2r^2 + p^2r^2 + p^2q^2) + 16(p^2 + q^2 + r^2) + 64.\]

By Vieta's we have $pqr = -3.$ Therefore, $p^2q^2r^2 = (-3)^2 = 9$. We use Newton's Sums to quickly compute $p^2 + q^2 + r^2:$

\[S_2 + 2S_1 + 2(-1) = 0\]

By Vieta's again, $S_1 = p + q + r -2$ which means $S_2 = p^2 + q^2 + r^2 = 6.$

Now all we need is $q^2r^2 + p^2r^2 + p^2q^2.$ Vieta's also tells us that $pq + qr + pr = -1,$ so we can take this equation and square both sides:

\[(pq + qr + pr)^2 = q^2r^2 + p^2r^2 + p^2q^2 + 2(qpr^2 + q^2rp + p^2rq) = 1.\]

We know $pqr = -3$ so we can substitute this in:

\[(pq + qr + pr)^2 = q^2r^2 + p^2r^2 + p^2q^2 -6(r + p + q) = 1.\]

We also know $p + q + r = -2$, which means $q^2r^2 + p^2r^2 + p^2q^2 = -11.$


So the answer is just $9 + 4(-11) + 16(6) + 64 = \boxed{125}.$

~grogg007

Solution 5 (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 6 (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 7

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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