Quadratics problem I came up with

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Quadratics problem I came up with

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V0305

#1 h Nov 3, 2024, 11:10 PM

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$P(x)$ is a quadratic polynomial such that $P(1) = 2$ and $P(2) = 5$ . Find all possible values of $P(7) - P(-4)$ .

This post has been edited 2 times. Last edited by V0305, Nov 3, 2024, 11:26 PM

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#2 h Nov 4, 2024, 1:48 AM • 2 Y

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We represent $P(x)$ as $ax^2+bx+c$ . From the first equation, we have $a+b+c=2$ and from the second we have $4a+2b+c=5$ . Note we are trying to find, $49a+7b-16a+4b-c$ which simplifies to $33a+11b$ . We also have from the first two equations $3a+b=3$ , and multiplying on both sides by $11$ we have $33a+11b=\boxed{33}$ , which is our answer.

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Dixit1

#3 h Apr 4, 2025, 4:56 AM

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Since p(1)=2 and p(2)=5, we have the equations

a+b+c=2 and 4a+2b+c=5
Equating both, we will get
b=3(1-a) and c=2a-1
Then p(7) would be 30a+20 and p(-4) would be 30a-13. Subtracting both the equations, we get the answer to be 33.

Hence the answer is 33

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#4 h Apr 5, 2025, 12:59 AM

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Sol

Also hi V0305!

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#5 h 3 hours ago

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Let P(x)= ax²+bx+c
Then, P(1)=a+b+c =2 ....(i)
P(2)=4a+2b+c=5....(ii)
Now (ii)-(i) gives 3a+b =3....(iii)

Now P(7)-P(-4)=11b+33a=11(b+3a) = 33 (from (iii))
Done !!

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