Solve 9x^2+3x-1

$$9 { x }^{ 2 } +3x-1$$

Factor

$$9\left(x-\frac{-\sqrt{5}-1}{6}\right)\left(x-\frac{\sqrt{5}-1}{6}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$9x^{2}+3x-1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-3±\sqrt{3^{2}-4\times 9\left(-1\right)}}{2\times 9}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 9\left(-1\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-3±\sqrt{9-36\left(-1\right)}}{2\times 9}$$

Multiply -36 times -1.

$$x=\frac{-3±\sqrt{9+36}}{2\times 9}$$

Add 9 to 36.

$$x=\frac{-3±\sqrt{45}}{2\times 9}$$

Take the square root of 45.

$$x=\frac{-3±3\sqrt{5}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-3±3\sqrt{5}}{18}$$

Now solve the equation x=\frac{-3±3\sqrt{5}}{18} when ± is plus. Add -3 to 3\sqrt{5}.

$$x=\frac{3\sqrt{5}-3}{18}$$

Divide -3+3\sqrt{5} by 18.

$$x=\frac{\sqrt{5}-1}{6}$$

Now solve the equation x=\frac{-3±3\sqrt{5}}{18} when ± is minus. Subtract 3\sqrt{5} from -3.

$$x=\frac{-3\sqrt{5}-3}{18}$$

Divide -3-3\sqrt{5} by 18.

$$x=\frac{-\sqrt{5}-1}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{5}}{6} for x_{1} and \frac{-1-\sqrt{5}}{6} for x_{2}.

$$9x^{2}+3x-1=9\left(x-\frac{\sqrt{5}-1}{6}\right)\left(x-\frac{-\sqrt{5}-1}{6}\right)$$

Evaluate

$$9x^{2}+3x-1$$