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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...

$a+b^2\ =\ \Big(\frac38\Big)+\Big(\frac{\sqrt{183}}{8}\Big)^2\ =\ \frac38+\frac{183}{64}\ =\ \frac{24}{64}+\frac{183}{64}\ =\ \frac{207}{64}$_