Let $a$ and $b$ be the roots of the quadratic equation $x^2-25x+80=-28x+75$. Compute $\frac{a^2}{b} + \frac{b^2}{a}$.
1 answer
We rewrite the given equation as $x^2 - 3x = -5$. From this, we see that $a^2 - 3a = -5$ and $b^2 - 3b = -5$. Multiplying these equations, we get $a^2 b^2 - 3ab = -5(-5) = 25$, so \[(ab)^2 - 3ab = 25.\]The given condition says $-3a = -3b = -3$, so \[\frac{a^2}{b} + \frac{b^2}{a} = a^3 \cdot \frac{1}{a} \cdot \frac{1}{b} + b^3 \cdot \frac{1}{b} \cdot \frac{1}{a} = ab( a^2 \cdot \frac{1}{ab} + b^2 \cdot \frac{1}{ab}) = ab(\frac{a^2 + b^2}{ab}) \]Since $a$ and $b$ are the roots of $x^2 -3x +5 = 0$, note by Vieta's formulas, $a+b=3$ and $ab=5$. Thus, $a^2 + b^2 = (a + b)^2 -2ab = 9 -2(5) = -1$, so $ab(\frac{a^2 + b^2}{ab}) = 5 \cdot -1 = \boxed{-5}.$