GRE problem for solving for X in x squared minus x minus 1 equals zero

alextrainer

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Solve equation for x

x squared minus x minus 1 equals zero

answer:
two solutions
1 plus square root of 5 divided by 2
and
1 minus square root of 5 divided by 2

I thought it was (x-1) (x plus 1) so 1, -1
 
Solve equation for x

x squared minus x minus 1 equals zero

answer:
two solutions
1 plus square root of 5 divided by 2
and
1 minus square root of 5 divided by 2

I thought it was (x-1) (x plus 1) so 1, -1

x2 - x - 1 = 0

Ax2 + Bx + C = 0

[x + {B + √(B2 - 4AC)}/(2A)] * [x + {B - √(B2 - 4AC)}/(2A)]

Your equation is:

x2 - x - 1 = 0

So you have:

A = 1 ; B = - 1 & C = -1 → √(B2 - 4AC) = √5

then the solutions of your equation are:

x1,2 = - {B ± √(B2 - 4AC)}/(2A) = 1/2 ± (√5)/2
 
Last edited by a moderator:
Solve equation for x
x squared minus x minus 1 equals zero
answer:
two solutions
1 plus square root of 5 divided by 2
and
1 minus square root of 5 divided by 2
I thought it was (x-1) (x plus 1) so 1, -1
(x1)(x+1)=x21\displaystyle (x-1)(x+1)=x^2-1.
 
Last edited by a moderator:
Thanks for your help. Why did you use the quadratic equation versus factoring (x-1)(x plus1)? How do I know whether to use factoring or quadratic equation?


x2 - x - 1 = 0

Ax2 + Bx + C = 0

[x + {B + √(B2 - 4AC)}/(2A)] * [x + {B - √(B2 - 4AC)}/(2A)]

Your equation is:

x2 - x - 1 = 0

So you have:

A = 1 ; B = - 1 & C = -1 → √(B2 - 4AC) = √5

then the solutions of your equation are:

x1,2 = - {B ± √(B2 - 4AC)}/(2A) = 1/2 ± (√5)/2
 
Thanks for your help. Why did you use the quadratic equation versus factoring (x-1)(x plus1)? How do I know whether to use factoring or quadratic equation?
Because f(x)=x2 - x - 1 cannot be factored easily.Thus quadratic equation is much quicker.
 
Thanks for your help. Why did you use the quadratic equation versus factoring (x-1)(x plus1)? How do I know whether to use factoring or quadratic equation?
Because, as PKA said before, x2x1\displaystyle x^2- x- 1 does not factor as "(x- 1)(x+ 1)". Multiplying (x- 1)(x+ 1) gives x21\displaystyle x^2- 1, not x2x1\displaystyle x^2- x- 1.

Personally, I would "complete the square: x2x1=x2x+14141=(x12)54\displaystyle x^2- x-1= x^2- x+ \frac{1}{4}- \frac{1}{4}- 1= (x- \frac{1}{2})- \frac{5}{4} which is of the form "a2b2\displaystyle a^2- b^2" and factors as (x1252)(x12+52)\displaystyle (x- \frac{1}{2}- \frac{\sqrt{5}}{2})(x- \frac{1}{2}+ \frac{\sqrt{5}}{2}) so that x=152\displaystyle x= \frac{1- \sqrt{5}}{2} and x=1+52\displaystyle x= \frac{1+ \sqrt{5}}{2}.
 
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