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

[alextrainer]

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

 

[Deleted member 4993]

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:

x_1,2 = - {B ± √(B² - 4AC)}/(2A) = 1/2 ± (√5)/2

 

[pka]

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

\(\displaystyle (x-1)(x+1)=x^2-1\).

 

[alextrainer]

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:

x_1,2 = - {B ± √(B² - 4AC)}/(2A) = 1/2 ± (√5)/2

 

[Deleted member 4993]

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.

 

[HallsofIvy]

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, \(\displaystyle x^2- x- 1\) does not factor as "(x- 1)(x+ 1)". Multiplying (x- 1)(x+ 1) gives \(\displaystyle x^2- 1\), not \(\displaystyle x^2- x- 1\).

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