solve x^2+x-1=0?

Question

quadratic equation

Answer 1

Here's the quadratic formula:
x=-b±√b^2-4ac/2a
a=1, b=1, c=-1
x=(-1±√(1)^2-4(1*-1)/2(1)
x=(-1±√1+4)/2
x=(-1±√5)/2

Answer 2

i'd say, use your handy dandy TI-89 or, settle for the good old fashioned quadratic formula which states:

if ax^2 + bx + c = 0, then

x = [ -b +/- sqrt(b^2 - 4ac) ] / 2a

In your equation, a=1, b=1, and c= -1.

Thus: x = [ -1 +/- sqrt (1^2 - 4(1)(-1) ) ] / 2(1)

Simplifying:

x = [ sqrt(5) - 1 ] / 2 AND [ -sqrt(5) - 1] / 2

Answer 3

In equation: a x^2 + bx + c = 0 then:

x =( -b +/- sqrt (b^2 - 4ac) ) / 2a

x = -1 +/- sqrt (1+4) / 2

x = ( -1 - sqrt (5)) / 2 and ( -1 + sqrt (5)) / 2

Answer 4

x^2+x-1=0

Use the formula where a=1, b=1, c=-1

x=0.618 or x=-1.618

I checked them using the calculator! ;)

Answer 5

x = -.5 ± (√5)/2

In decimal form this is .618..... and 1.618...., the ratio of the sides of the 'golden rectangle' known to the ancient Greeks and incorporated into much of their architecture.

Answer 6

ax^2+bx+c=0

x= ( -b +/- sqrt(b^2-4ac) ) / (2a)

in your case, a=1, b=1, c=-1.

Answer 7

use quadratic formula...or factor it out, use your graphing calculator
(if you have one)

Answer 8

its not solvable or the answer is in fractions decimals...

Answer 9

x=[-1+/-sq.rt(1+4)]/2
=[-1+/-sq.rt5]/2
x1=(-1+rt5)/2
x2=(-1-rt5)/2

Answer 10

DO. YOUR. OWN. HOMEWORK.