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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.

 Jan 9, 2019
 #1
avatar+103138 
+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

 

 

cool cool cool

 Jan 9, 2019
 #2
avatar+18965 
+1

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

 Jan 9, 2019

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