Let $a$ and $b$ be the roots of the quadratic equation $x^2-25x+80=-28x+75$. Compute $\frac{a^2}{b} + \frac{b^2}{a}$.
Simplify:
\(x^2+3x+5=0\)
Plugging in the values a = 1, b = 3, and c = 5 into the quadratic equation \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), we get:
\(x = {-3 \pm \sqrt{9-20} \over 2}\)
The discriminant, or \(\sqrt{b^2-4ac}\), is less than 0, meaning that this quadratic equation has complex roots, which include imaginary numbers, as we are dealing with the square root of a negative number.
Further simplifying, we get:
\(x = {-3 \pm \sqrt{11} i \over 2}\), where i is not in the square root, but rather it is the square root of 11 multiplied by i.
We get that the complex roots of the quadratic equation are:
\(x = \frac{-\sqrt{11}i-3}{2}\), \(x = \frac{\sqrt{11}i-3}{2}\)
So, we get that a = \( \frac{-\sqrt{11}i-3}{2}\), and b = \(\frac{\sqrt{11}i-3}{2}\).
Plugging in these values into the expression \(\frac{a^2}{b} + \frac{b^2}{a}\), we get:
\(\frac{ (\frac{-\sqrt{11}i-3}{2})^2}{ \frac{\sqrt{11}i-3}{2}} + \frac{( \frac{\sqrt{11}i-3}{2})^2}{ \frac{-\sqrt{11}i-3}{2}}\)
Can you do the rest?
a^2 / b + b^2 / a = [ a^3 + b^3 ] / [ab ] ^2 = [(a + b) (a^2 + b^2 - ab ) ] / [ab]
x^2 -25x + 80 = -28x + 75 simplify
x^2 + 3x + 5 = 0
We have the form mx^2 + nx + c = 0
Sum of roots
a + b = -n/m = -3/1 = -3
Product of roots = ab = c / m = 5/1 = 5
a + b = -3 square both sides
a^2 + 2ab + b^2 = 9
a^2 +2(5) + b^2 = 9
a^2 + 10 + b^2 = 9
a^2 + b^2 = -1
So
[(a + b) (a^2 + b^2 - ab) ] / [ab] =
[ ( -3)( -1 - 5) ] / [5] =
[ 18 ] / [5 ] =
18 / 5 = 3.6