The quadratic \(2x^2-3x+27\) has two imaginary roots. What is the sum of the squares of these roots? Express your answer as a decimal rounded to the nearest hundredth.

Guest Jan 9, 2019

#1**+1 **

2x^2 - 3x + 27 = 0

Let the roots be p and q

The sum of these roots = 3/2

The product of these roots = 27/2

So

p + q = 3/2 (1)

And

pq = 27/2

2 pq = 27 (2)

Square both sides of (1)

p^2 + 2pq + q^2 = 9/4 sub (2) into this

p^2 + (27) + q^2 = 9/4

p^2 + q^2 = 9 / 4 - 27

p^2 + q^2 = 9/4 - 108 / 4

p^2 + q^2 = -99 / 4

p^2 + q^2 = -24.75

CPhill Jan 9, 2019

#2**+2 **

Thanx, Chris..... here is another method:

Use quadratic formula to find the two roots as 3/4 +- 3.5968i

(3/4+ 3.5968i)^2 = 9/16 +3/2 (3.5968i) - 12.9375

(3/4- 3.5968i)^2 = 9/16 -3/2 (3.5968i) - 12.9375

Added together 9/8 - 25.875 = - 24.75

ElectricPavlov Jan 9, 2019