The solutions to $4x^2 + 3 = 3x - 9$ can be written in the form $x = a \pm b i,$ where $a$ and $b$ are real numbers. What is $a + b^2$? Express your answer as a fraction.
+9466 The solutions to \(4x^2 + 3 = 3x - 9\) can be written in the form \(x = a \pm b i,\) where \(a\) and \(b\) are real numbers. What is \(a + b^2\) ? Express your answer as a fraction.
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\(4x^2 + 3 = 3x - 9\)
Subtract 3x from both sides and add 9 to both sides of the equation.
\(4x^2-3x + 12 = 0\)
Now we can use the quadratic formula to solve for x
\(x\ = \ \frac{3\pm\sqrt{3^2-4(4)(12)}}{2(4)}\ =\ \frac{3\pm\sqrt{-183}}{8}\ =\ \frac{3\pm\sqrt{183}i}{8}\ =\ \frac38\pm\frac{\sqrt{183}}{8}i\)
And now we have the solutions in the form \(x = a \pm b i\) so we can see...
\(a+b^2\ =\ \Big(\frac38\Big)+\Big(\frac{\sqrt{183}}{8}\Big)^2\ =\ \frac38+\frac{183}{64}\ =\ \frac{24}{64}+\frac{183}{64}\ =\ \frac{207}{64}\)_
