Question

Solve the quadratic equation.
9x^2–2=2x

Answer 1

To solve the quadratic equation \(9x^2 - 2 = 2x\), we first rearrange the equation into standard form \(ax^2 + bx + c = 0\).

Starting with the original equation:

\[
9x^2 - 2 = 2x
\]

we subtract \(2x\) from both sides:

\[
9x^2 - 2 - 2x = 0
\]

Rearranging gives:

\[
9x^2 - 2x - 2 = 0
\]

Now we can identify the coefficients:

- \(a = 9\)
- \(b = -2\)
- \(c = -2\)

Next, we use the quadratic formula to find the roots:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Calculating the discriminant \(b^2 - 4ac\):

\[
b^2 - 4ac = (-2)^2 - 4(9)(-2) = 4 + 72 = 76
\]

Now substituting back into the quadratic formula:

\[
x = \frac{-(-2) \pm \sqrt{76}}{2(9)} = \frac{2 \pm \sqrt{76}}{18}
\]

Simplifying \(\sqrt{76}\):

\[
\sqrt{76} = \sqrt{4 \cdot 19} = 2\sqrt{19}
\]

Substituting this back into the equation gives:

\[
x = \frac{2 \pm 2\sqrt{19}}{18}
\]

Factoring out the common term:

\[
x = \frac{2(1 \pm \sqrt{19})}{18} = \frac{1 \pm \sqrt{19}}{9}
\]

Thus, the solutions to the equation \(9x^2 - 2 = 2x\) are:

\[
x = \frac{1 + \sqrt{19}}{9} \quad \text{and} \quad x = \frac{1 - \sqrt{19}}{9}
\]