Question

what are the roots of the equation?
$6x^2 - 7x + 2 = 0$
answer
$\frac{7\pm i\sqrt{97}}{12}$ $-\frac{1}{2}$ and $-\frac{2}{3}$
$\frac{2}{3}$ and $\frac{1}{2}$ $\frac{-7\pm i\sqrt{97}}{12}$

Explanation:

Step1: Factor the quadratic equation

We try to factor \(6x^{2}-7x + 2 = 0\). We need two numbers that multiply to \(6\times2 = 12\) and add up to \(- 7\). The numbers are \(-3\) and \(-4\).
So we rewrite the middle term:
\(6x^{2}-3x-4x + 2 = 0\)
Group the terms:
\((6x^{2}-3x)-(4x - 2)=0\)
Factor out the common factors from each group:
\(3x(2x - 1)-2(2x - 1)=0\)
Now we can factor out \((2x - 1)\):
\((3x - 2)(2x - 1)=0\)

Step2: Solve for \(x\)

Set each factor equal to zero:

• For \(3x - 2 = 0\), we add \(2\) to both sides: \(3x=2\), then divide by \(3\): \(x=\frac{2}{3}\)
• For \(2x - 1 = 0\), we add \(1\) to both sides: \(2x = 1\), then divide by \(2\): \(x=\frac{1}{2}\)

Answer:

\(\frac{2}{3}\) and \(\frac{1}{2}\)