Question

two solutions for all quadratic equations quick check
solve the quadratic equation $6x^2 - 3x + 6 = 0$ which of the following expresses its solutions in the form $a \pm bi$?
(1 point)
$\circ \\ \frac{1}{4} \pm \frac{\sqrt{15}}{4}i$
$\circ \\ \frac{1}{2} \pm \frac{\sqrt{15}}{2}i$
$\circ \\ -\frac{1}{4} \pm \frac{\sqrt{15}}{4}i$
$\circ \\ \frac{1}{4} \pm \frac{\sqrt{17}}{4}i$

Explanation:

Step1: Recall Quadratic Formula

The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For the equation \(6x^2-3x + 6 = 0\), we have \(a = 6\), \(b=-3\), and \(c = 6\).

Step2: Calculate the Discriminant

First, calculate the discriminant \(D=b^2-4ac\). Substitute the values: \(D=(-3)^2-4\times6\times6=9 - 144=- 135\). Wait, no, wait, let's recalculate: \(b=-3\), so \(b^2 = (-3)^2=9\), \(4ac=4\times6\times6 = 144\), so \(D=9-144=-135\)? Wait, no, the options have \(\sqrt{15}\) or \(\sqrt{17}\), maybe I made a mistake. Wait, the equation is \(6x^2-3x + 6 = 0\), divide the entire equation by 3: \(2x^2 - x+2 = 0\). Now \(a = 2\), \(b=-1\), \(c = 2\). Then discriminant \(D=(-1)^2-4\times2\times2=1 - 16=-15\). Ah, that's better. So \(D=-15\), so \(\sqrt{D}=\sqrt{-15}=i\sqrt{15}\).

Step3: Apply Quadratic Formula

Now, using the quadratic formula \(x=\frac{-b\pm\sqrt{D}}{2a}\), with \(a = 2\), \(b=-1\), \(D=-15\). So \(x=\frac{-(-1)\pm i\sqrt{15}}{2\times2}=\frac{1\pm i\sqrt{15}}{4}=\frac{1}{4}\pm\frac{\sqrt{15}}{4}i\). Wait, but let's check the original equation again. Wait, the original equation is \(6x^2-3x + 6 = 0\), so \(a = 6\), \(b=-3\), \(c = 6\). Then \(x=\frac{-(-3)\pm\sqrt{(-3)^2-4\times6\times6}}{2\times6}=\frac{3\pm\sqrt{9 - 144}}{12}=\frac{3\pm\sqrt{-135}}{12}=\frac{3\pm3i\sqrt{15}}{12}=\frac{1\pm i\sqrt{15}}{4}=\frac{1}{4}\pm\frac{\sqrt{15}}{4}i\). Yes, that's correct. So the solution is \(\frac{1}{4}\pm\frac{\sqrt{15}}{4}i\), which is the first option.

Answer:

\(\boldsymbol{\frac{1}{4}\pm\frac{\sqrt{15}}{4}i}\) (the first option: \(\frac{1}{4}\pm\frac{\sqrt{15}}{4}i\))