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Let $a$ and $b$ be the roots of the quadratic equation $x^2-25x+80=-28x+75$. Compute $\frac{a^2}{b} + \frac{b^2}{a}$.

Jan 14, 2024

#1
+289
+1

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?

Jan 14, 2024
#2
+128732
+1

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

Jan 14, 2024