Problem 2 · AMC 8 Stretch
Stretch
Algebra & Patterns
Number Theory
sum-and-product-of-rootstest-divisors
In the equation \(Ax^2+Bx+C=0\), the two solutions (roots) turn out to be exactly \(A\) and \(B\). Here \(A\), \(B\), \(C\) are nonzero whole numbers (they may be negative). Find \(A\), \(B\), and \(C\).
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Answer: A=-2, B=4, C=16
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Hint 1 of 4
Useful fact: for \(Ax^2+Bx+C=0\), the two roots add up to \(-\tfrac{B}{A}\) and multiply to \(\tfrac{C}{A}\). Here the roots are \(A\) and \(B\) themselves, so \(A+B=-\tfrac{B}{A}\).
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Hint 2 of 4
Multiply that by \(A\) to clear the fraction: \(A^2+AB=-B\). Get \(B\) alone: \(B(A+1)=-A^2\), so \(B=\dfrac{-A^2}{A+1}\).
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Hint 3 of 4
For \(B\) to be a whole number, \(A+1\) has to divide \(A^2\) evenly. A short division shows the only way is \(A+1=-1\), giving \(A=-2\). Plug in to find \(B\).
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Approach: Sum of roots equals -B/A, force the leftover to be a whole number
- For \(Ax^2+Bx+C=0\) the roots add to \(-\tfrac{B}{A}\). Since the roots are \(A\) and \(B\), \(A+B=-\tfrac{B}{A}\).
- Multiply both sides by \(A\): \(A^2+AB=-B\), so \(B(A+1)=-A^2\) and \(B=\dfrac{-A^2}{A+1}\).
- Rewrite \(\dfrac{A^2}{A+1}=A-1+\dfrac{1}{A+1}\); the leftover \(\dfrac{1}{A+1}\) is a whole number only when \(A+1=\pm1\). Since \(A\ne0\), we can't use \(A+1=1\), so \(A+1=-1\), giving \(A=-2\).
- Then \(B=\dfrac{-(-2)^2}{-2+1}=\dfrac{-4}{-1}=4\).
- Use that \(B=4\) is a root of \(-2x^2+4x+C=0\): plug \(x=4\) to get \(-2(16)+4(4)+C=0\Rightarrow-32+16+C=0\Rightarrow C=16\).
- Check: \(-2x^2+4x+16=-2(x-4)(x+2)\), whose roots are \(4\) and \(-2\) — exactly \(B\) and \(A\). So \(A=-2,\ B=4,\ C=16\).
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