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.
+130071 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
+37157 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
