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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.

 Dec 27, 2020
 #1
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Answered here:

 

https://web2.0calc.com/questions/help_30050

 Dec 27, 2020
 #2
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Solution #1: \(2x^2 - 3x + 27 = 0 \)        

 

Let the roots be \(p\) and \(q\).

 

The sum of these roots \(=\frac32\).

The product of these roots \(=\frac{27}{2}\).

 

So

\(p + q   =  \frac32\) (1)

And

\(pq =\frac{27}{2}\)

\(2 pq =  27\) (2)

 

Square both sides of (1)

 

\(p^2 + 2pq  + q^2  = \frac94\) Sub (2) into this

 

\(p^2 + (27) + q^2 = \frac94\)


\(p^2 + q^2 =      \frac94 - 27 \)

 

\(p^2 + q^2 =   \frac94 - \frac{108}{4}\)

 

\(p^2 + q^2 =  -\frac{99}{4}\)

 

\(p^2 + q^2 =  \boxed{-24.75}\)

 

Solution #2: Use quadratic formula to find the two roots as  \(\frac34 + - 3.5968i\).

 

\((\frac34+ 3.5968i)^2 = \frac{9}{16} +\frac32 (3.5968i) - 12.9375\)

\((\frac34- 3.5968i)^2 = \frac{9}{16} -\frac32 (3.5968i) - 12.9375\)

 

Added together, \(\frac98 - 25.875 = \boxed{-24.75}\).

 Dec 28, 2020

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