How do you solve $x^{2} - x - 1 = 0$ $x^2-x-1=0$ by completing the square?

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Answer 1 · 2018-05-24T11:45:42

We have

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

$x^{2} - 2 \cdot \frac{1}{2} \cdot x + {(\frac{1}{2})}^{2} - \frac{5}{4} = 0$ $x^2-2*1/2*x+(1/2)^2-5/4=0$

${(x - \frac{1}{2})}^{2} - \frac{5}{4} = 0$ $(x-1/2)^2-5/4=0$

${(x - \frac{1}{2})}^{2} = \frac{5}{4}$ $(x-1/2)^2=5/4$

Taking square roots in both sides

$∣ ∣ ∣ x - \frac{1}{2} ∣ ∣ ∣ = \frac{\sqrt{5}}{2}$ $|x-1/2|=sqrt5/2$

and

$(x - \frac{1}{2}) = \pm \frac{\sqrt{5}}{2}$ $(x-1/2)=+- sqrt5/2$

Hence the two solutions are

$x_{1} = \frac{1}{2} + \frac{\sqrt{5}}{2}$ $x_1=1/2+sqrt5/2$ and $x_{2} = \frac{1}{2} - \frac{\sqrt{5}}{2}$ $x_2=1/2-sqrt5/2$

How do you solve x2−x−1=0x^2-x-1=0 by completing the square?