+0

# Help

0
211
1

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.

Jun 25, 2019

#1
+8842
+4

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.

----------

$$4x^2 + 3 = 3x - 9$$

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...

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

Jun 25, 2019