Solve x+1*x*x-1

$$x+1 \times x \times x-1$$

Evaluate

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

Solution Steps

Multiply x and x to get x^{2}.

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

For any term t, t\times 1=t and 1t=t.

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

Factor

$$\left(x-\frac{-\sqrt{5}-1}{2}\right)\left(x-\frac{\sqrt{5}-1}{2}\right)$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$factor(x+1x^{2}-1)$$

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.

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

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.

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

Square 1.

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

Multiply -4 times -1.

$$x=\frac{-1±\sqrt{1+4}}{2}$$

Add 1 to 4.

$$x=\frac{-1±\sqrt{5}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{5}}{2} when ± is plus. Add -1 to \sqrt{5}.

$$x=\frac{\sqrt{5}-1}{2}$$

Now solve the equation x=\frac{-1±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from -1.

$$x=\frac{-\sqrt{5}-1}{2}$$

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}}{2} for x_{1} and \frac{-1-\sqrt{5}}{2} for x_{2}.

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