Question

complex numbers & higher order polynomials quick check
which answer shows the conjugate pair solutions to the equation $x^2 - 3x + 8 = -2$? (1 point)
$\frac{3}{2} \pm \frac{\sqrt{15}}{2}i$
$-\frac{3}{2} \pm \frac{\sqrt{31}}{2}i$
$\frac{3}{2} \pm \frac{\sqrt{31}}{2}i$
$\frac{3}{2} \pm \frac{\sqrt{23}}{2}i$

Explanation:

Step1: Rewrite the equation

First, rewrite the given equation \(x^{2}-3x + 8=-2\) in standard quadratic form \(ax^{2}+bx + c = 0\). Add 2 to both sides: \(x^{2}-3x+10 = 0\). Here, \(a = 1\), \(b=-3\), \(c = 10\).

Step2: Use quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 1\), \(b=-3\), \(c = 10\) into the formula. First, calculate the discriminant \(\Delta=b^{2}-4ac=(-3)^{2}-4\times1\times10=9 - 40=-31\). Then, \(x=\frac{-(-3)\pm\sqrt{-31}}{2\times1}=\frac{3\pm\sqrt{31}i}{2}=\frac{3}{2}\pm\frac{\sqrt{31}}{2}i\).

Answer:

\(\boldsymbol{\frac{3}{2}\pm\frac{\sqrt{31}}{2}i}\) (the third option, i.e., \(\frac{3}{2}\pm\frac{\sqrt{31}}{2}i\))