Of course, let's solve the equation [tex]\(-9x - 1 = -9x^2\)[/tex] step by step.
First, move all the terms to one side of the equation to set it equal to zero:
[tex]\[
-9x - 1 + 9x^2 = 0
\][/tex]
Rewriting the equation, we get:
[tex]\[
9x^2 - 9x - 1 = 0
\][/tex]
Next, we use the quadratic formula to find the roots of the equation. The quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For our equation, [tex]\(a = 9\)[/tex], [tex]\(b = -9\)[/tex], and [tex]\(c = -1\)[/tex].
Plugging in these values, we have:
[tex]\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 9 \cdot (-1)}}{2 \cdot 9}
\][/tex]
Simplify the expression inside the square root:
[tex]\[
b^2 - 4ac = (-9)^2 - 4 \cdot 9 \cdot (-1) = 81 + 36 = 117
\][/tex]
So the equation simplifies to:
[tex]\[
x = \frac{9 \pm \sqrt{117}}{18}
\][/tex]
Now, recall that [tex]\(\sqrt{117}\)[/tex] can be simplified as [tex]\(\sqrt{9 \cdot 13} = 3\sqrt{13}\)[/tex]. Therefore, we get:
[tex]\[
x = \frac{9 \pm 3\sqrt{13}}{18}
\][/tex]
Divide both the numerator terms by the common factor of 3:
[tex]\[
x = \frac{3 \pm \sqrt{13}}{6}
\][/tex]
Separating this into two solutions:
[tex]\[
x = \frac{3 + \sqrt{13}}{6} \quad \text{and} \quad x = \frac{3 - \sqrt{13}}{6}
\][/tex]
We can further simplify these solutions:
[tex]\[
x = \frac{1}{2} + \frac{\sqrt{13}}{6} \quad \text{and} \quad x = \frac{1}{2} - \frac{\sqrt{13}}{6}
\][/tex]
Hence, the solutions to the equation [tex]\(-9x - 1 = -9x^2\)[/tex] are:
[tex]\[
x = \frac{1}{2} \pm \frac{\sqrt{13}}{6}
\][/tex]
