Question

using the quadratic formula to solve $4x^2 - 3x + 9 = 2x + 1$, what are the values of $x$?
options:
$\frac{1 pm sqrt{153}i}{8}$
$\frac{5 pm sqrt{153}i}{8}$
$\frac{5 pm sqrt{103}i}{8}$
$\frac{1 pm sqrt{153}}{8}$

Explanation:

Step1: Rearrange the equation

First, we need to rewrite the given equation \(4x^{2}-3x + 9=2x + 1\) in standard quadratic form \(ax^{2}+bx + c = 0\).
Subtract \(2x\) and \(1\) from both sides:
\(4x^{2}-3x-2x + 9 - 1=0\)
Simplify the like - terms: \(4x^{2}-5x + 8 = 0\)

Step2: Identify \(a\), \(b\), and \(c\)

For a quadratic equation \(ax^{2}+bx + c = 0\), in the equation \(4x^{2}-5x + 8 = 0\), we have \(a = 4\), \(b=-5\), and \(c = 8\).

Step3: Use the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac\).
Substitute \(a = 4\), \(b=-5\), and \(c = 8\) into the discriminant formula:
\(\Delta=(-5)^{2}-4\times4\times8=25 - 128=- 103\)? Wait, no, wait:
Wait, let's recalculate the discriminant. Wait, \(b=-5\), so \(b^{2}=(-5)^{2}=25\), \(4ac = 4\times4\times8=128\), so \(\Delta=25-128=- 103\)? Wait, no, the original equation after rearrangement: Wait, the original equation is \(4x^{2}-3x + 9=2x + 1\). Let's do the rearrangement again:
\(4x^{2}-3x-2x+9 - 1=4x^{2}-5x + 8 = 0\). So \(a = 4\), \(b=-5\), \(c = 8\). Then \(\Delta=b^{2}-4ac=(-5)^{2}-4\times4\times8=25 - 128=-103\)? But the options have \(\sqrt{153}\) or \(\sqrt{103}\). Wait, I must have made a mistake in rearrangement.
Wait, original equation: \(4x^{2}-3x + 9=2x + 1\)
Subtract \(2x\) and \(1\) from both sides: \(4x^{2}-3x-2x+9 - 1=4x^{2}-5x + 8 = 0\). Wait, maybe I made a mistake in the sign of \(b\). Wait, the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(b\) is the coefficient of \(x\). In \(4x^{2}-5x + 8 = 0\), \(b=-5\), so \(-b = 5\).
Now, \(b^{2}-4ac=(-5)^{2}-4\times4\times8=25 - 128=-103\)? No, that's not matching the options. Wait, maybe I rearranged the equation wrong. Let's start over.
Original equation: \(4x^{2}-3x + 9=2x + 1\)
Bring all terms to the left - hand side: \(4x^{2}-3x-2x + 9 - 1=4x^{2}-5x + 8 = 0\). Wait, maybe the original equation was \(4x^{2}-3x + 9=2x + 1\), let's check the discriminant again. Wait, maybe I misread the equation. Let's check the options. The options have \(\sqrt{153}\) or \(\sqrt{103}\). Let's recalculate the discriminant with a different approach.
Wait, maybe the equation is \(4x^{2}-3x + 9=2x + 1\), so \(4x^{2}-3x-2x+9 - 1=4x^{2}-5x + 8 = 0\). Then \(a = 4\), \(b=-5\), \(c = 8\). Then \(x=\frac{5\pm\sqrt{(-5)^{2}-4\times4\times8}}{2\times4}=\frac{5\pm\sqrt{25 - 128}}{8}=\frac{5\pm\sqrt{-103}}{8}=\frac{5\pm\sqrt{103}i}{8}\). Ah! Because \(\sqrt{-103}=\sqrt{103}\times\sqrt{-1}=\sqrt{103}i\) (since \(i^{2}=-1\)).

Answer:

\(\frac{5\pm\sqrt{103}i}{8}\) (the third option: \(\frac{5\pm\sqrt{103}i}{8}\))