+303 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?
+16 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.
+16 Had there not been any limitations; we could've done the following:
First, we can start with some basic rearrangements.
\(2x^2 - 10x + 13 = -6x^2 - 18x - 15\),
\(8x^2 + 8x + 28 = 0\),
\(x^2 + x + \frac{7}{2} = 0 \)
.
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 \(a\cdot b = -\frac{1}{2}\cdot\frac{\pm\sqrt{13}}{2} = \frac{\pm\sqrt{13}}{4}.\)
