SOLUTION: x^2-x-1=0

Question 29997:

x^2-x-1=0

Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
I need help solving these two problems.
2x^2-3x+1=0 ----(1)
2x^2-2x-x+1 = 0
(2x^2-2x)-x+1 = 0
2x(x-1)-1(x-1) = 0
2xp-p = 0
p(2x-1)= 0 where p= (x-1)
(x-1)(2x-1) = 0
(x-1)=0 gives x=1
(2x-1)=0 gives 2x = 1 implies x = 1/2
Answer: x = 1 and x = 1/2
Verification: x = 1 in (1) implies LHS = 2x^2-3x+1 = 2-3+1 = 0 = RHS
x=1/2 in (1) implies LHS = 2x^2-3x+1 = 2X(1/4)-3X(1/2)+1 = 1/2-3/2+1 =0 =RHS
Note: How did the middle term (-3x)get expressed
as the sum of (-2x) and (-x) and why?
Find the product of the square term and the constant term in (1)
That is (2x^2)X(1) = 2x^2.
Observe the middle term. It is negative and the above product is positive. So you conclude that the two parts into which the mid term should be split must be both negative so that negative added to negative is negative and negative multiplied by negative is positive. That decided you split the produtct without leaving any factor into two parts and attach negative sign to both the parts.
Here (2x^2) = (-2x)X(-x)
2)x^2-x-1=0 ----(2)
Multiply by 4
4x^2-4x-4 = 0
4x^2-4x = 4
(2x)^2-2X(2x)X1 = 4
(the LHS resembles a^2-2ab and so we need b^2 to complete the square.)
Here a = 2x and b= 1
Adding 1 to both the sides
(2x)^2-2X(2x)X1 + 1^2 = 4 +1^2
(2x-1)^2 = 5
Taking sqrt
(2x-1) = +sqrt(5) and (2x-1) = -sqrt(5)
That is 2x = 1+sqrt(5) and 2x = 1-sqrt(5)
Answer: x= [1+sqrt(5)]/2 and x = [1-sqrt(5)]/2
Verification:Putting x= [1+sqrt(5)]/2 in (2)
LHS = {[1+sqrt(5)]/2 }^2-[1+sqrt(5)]/2 -1
=(1/4)[(1+5+2rt5)]-[1+sqrt(5)]/2 -1
=(1/4)[(1+5+2rt5)-2(1+rt5)-4]
=(1/4)X[6+2rt5-2-2rt5-4]
=(1/4)X(0) = 0 = RHS
Since surd roots occur in conjugate pairs,
there is no need to verify the second value.
Note: why multiply by 4 instead of directly proceeding
with perfecting the square?
To avoid fractions while converting into a perfect square

RELATED QUESTIONS
x^2+1=0 (answered by Nate)

1/x-2 <=... (answered by Fombitz)

1/x-2>0 (answered by lynnlo)

x^2-x-1=0 (answered by jsmallt9)

X^2+x+1=0 (answered by solver91311)

X^2+x+1=0 (answered by harpazo)

X^2 + x +1 =... (answered by KMST,Alan3354)

x^2+x+1=0... (answered by stanbon)