Problem 25 · 2016 Math Kangaroo
Stretch
Algebra & Patterns
substitutiondifference-of-squares
The equations \(x^2 + ax + b = 0\) and \(x^2 + bx + a = 0\) both have real solutions. It is known that the sum of the squares of the solutions of the first equation is equal to the sum of the squares of the solutions of the second equation, and that \(a \ne b\). Then \(a + b\) equals
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Answer: B — -2
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Hint 1 of 3
By Vieta, the sum of the squares of the roots is \((\text{sum})^2 - 2(\text{product})\).
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Hint 2 of 3
Set the two sums-of-squares equal, then factor the resulting symmetric equation.
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Hint 3 of 3
The condition \(a \ne b\) lets you cancel one factor.
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Approach: Vieta plus factoring
- For \(x^2+ax+b\) the roots have sum \(-a\) and product \(b\), so their squares sum to \(a^2 - 2b\); for \(x^2+bx+a\) it is \(b^2 - 2a\).
- Setting \(a^2 - 2b = b^2 - 2a\) gives \(a^2 - b^2 + 2a - 2b = 0\), i.e. \((a-b)(a+b+2) = 0\).
- Since \(a \ne b\), the other factor vanishes: \(a + b = -2\), answer B.
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