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.
+122 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, \(\frac98 - 25.875 = \boxed{-24.75}\).
