How do you solve $x^{2} - x - 1 = 0$ $x^2-x-1=0$ using the quadratic formula?

Question

116529 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-24T07:22:41

See below

$x = 1.62 or x = - 0.62$ $x=1.62 or x= -0.62$

The quadratic formula is $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $[-b+-sqrt(b^2-4ac)]/(2a)$

this is used for formulas $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

in your problem, $x^{2} - x - 1 = 0$ $" "x^2−x−1=0" "$ a=1, b=-1, c=-1#

substitute into quadratic $\frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (- 1)}}{2 \cdot 1}$ $" "[-(-1)+-sqrt((-1)^2-4(1)(-1))]/(2*1)$

this equals $\frac{1 \pm \sqrt{1 + 4}}{2}$ $" "(1+-sqrt(1+4))/2$

$\frac{1 + \sqrt{5}}{2} or \frac{1 - \sqrt{5}}{2}$ $(1+sqrt5)/2 or (1-sqrt5)/2$

$x = 1.62 or x = - 0.62$ $x=1.62 or x= -0.62$

How do you solve x2−x−1=0x^2-x-1=0 using the quadratic formula?