Let $a$ and $b$ be the roots of the quadratic equation $x^2-25x+80=-28x+75$. Compute $\frac{a^2}{b} + \frac{b^2}{a}$.
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 $ab(\frac{a^2 + b^2}{ab}) = 5 \cdot -1 = \boxed{-5}.$
To find the roots of the quadratic equation $x^2-25x+80=-28x+75$, we need to simplify and solve for $x$.
First, we combine like terms:
$x^2 - 25x + 80 = -28x + 75$
Next, we move all terms to one side of the equation:
$x^2 - 25x + 28x + 80 - 75 = 0$
Simplifying further:
$x^2 + 3x + 5 = 0$
Now, we can use the quadratic formula to find the roots of this equation:
$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Comparing this to our quadratic equation, we have $a=1$, $b=3$, and $c=5$.
Substituting these values into the quadratic formula:
$x = \frac{-3\pm\sqrt{3^2-4(1)(5)}}{2(1)}$
Simplifying:
$x = \frac{-3\pm\sqrt{9-20}}{2}$
$x = \frac{-3\pm\sqrt{-11}}{2}$
Since the discriminant, $b^2-4ac$, is negative, the roots of the equation are complex. Let's let $a$ and $b$ represent the two complex roots.
Now, to compute $\frac{a^2}{b} + \frac{b^2}{a}$, we substitute the values of $a$ and $b$:
$\frac{a^2}{b} + \frac{b^2}{a} = \frac{(a^2)^2}{ab} + \frac{(b^2)^2}{ab}$
$\frac{a^2}{b} + \frac{b^2}{a} = \frac{a^4 + b^4}{ab}$
Since we're dealing with complex roots, we don't know the exact values of $a$ and $b$. Therefore, we cannot simplify $\frac{a^4 + b^4}{ab}$ any further without knowing the specific values of $a$ and $b$.