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# Algebra

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The solutions to
2x^2 - 10x + 13 = -6x^2 - 18x - 15
are a+bi and a-bi, where a and b are positive. What is a\cdot b?

Jul 25, 2024

#1
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The value for a can't be positive for the solutions for that equation. Since the roots are literally $$\frac{-1\pm\sqrt{13}i}{2}$$. b > 0 but not a.

Jul 25, 2024
#2
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Had there not been any limitations; we could've done the following:

$$2x^2 - 10x + 13 = -6x^2 - 18x - 15$$,

$$8x^2 + 8x + 28 = 0$$

$$x^2 + x + \frac{7}{2} = 0$$

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We turn this into a monic quadratic to simplify our working out. Since the roots of the quadratic are $$a+bi, a-bi$$

, we can use Vieta's formula to see that:

$$(a+bi)+(a-bi) = 2a = -1 \rightarrow a = -\frac{1}{2}$$
$$(a+bi)\cdot(a-bi) = a^2 + b^2 = \frac{7}{2}$$

We can substitute our value for a and get $$a^2 + b^2 = \frac{7}{2} \rightarrow (-\frac{1}{2})^2 + b^2 = \frac{7}{2} \rightarrow b^2 = \frac{13}{4} \therefore b = \frac{\pm\sqrt{13}}{2}.$$

Which means $$a\cdot b = -\frac{1}{2}\cdot\frac{\pm\sqrt{13}}{2} = \frac{\pm\sqrt{13}}{4}.$$

Jul 25, 2024