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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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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