Question

Find the real solutions of$x^2-x-1=0$

Short answer

The solution set is$\left\{\frac{1+\sqrt{5}}{-2},\frac{1-\sqrt{5}}{-2}\right\}$.

Step-by-step solution

Step 1. Given Information

We have to find the real solutions of$x^2-x-1=0$

Step 2. The equation is in standard form, so we compare it to ax2+bx+c=0 to find a,b and c.

$$\begin{gathered} x^2-x-1=0 \\ a{x}^2+bx+c=0 \end{gathered}$$

So$a=1,b=-1,c=-1$

Step 3. With a=1, b=-1 and c=-1, we evaluate the discriminant b2-4ac.

$$\begin{gathered} =b^2-4ac \\ ={(-1)}^2-4\times 1\times (-1) \\ =1+4 \\ =5 \end{gathered}$$

Since $b^2-4ac\ge 0$, there are two real solutions.

Step 4. We use the quadratic formula witha=1, b=-1, c=-1 and b2-4ac=5.

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-1)\pm \sqrt{5}}{2\times (-1)} \\ x=\frac{1\pm \sqrt{5}}{-2} \end{gathered}$$

The solution set is$\left\{\frac{1+\sqrt{5}}{-2},\frac{1-\sqrt{5}}{-2}\right\}$