How do you solve using the quadratic formula #2x^2 - 2x = 1#?

2 Answers
Apr 29, 2018

Answer:

#x_1=(1+sqrt(3))/2 or x_2=(1-sqrt(3))/2#

Explanation:

#2x^2-2x=1|-1#
#2x^2-2x-1=0|:2#
#x^2-x-1/2=0#
#x_(1,2)=-(p/2)+-sqrt((p/2)^2-q)#
#x_(1,2)=-(-1/2)+-sqrt((-1/2)^2-(-1/2))#
#x_(1,2)=1/2+-sqrt(1/4+1/2)=1/2+-sqrt(3)/2#

Apr 29, 2018

Answer:

#x=(1pmsqrt3)/2#

Explanation:

Minus #1#:

#2x^2-2x-1=0#

Use the formula:

#rArr x=-bpm(sqrt(b^2-4ac))/(2a)#

Use the quadratic to find the values:

#2x^2-2x-1#

#a=2#

#b=-2#

#c=-1#

#therefore# #rArr x=-(-2)pm(sqrt((-2)^2-4(2)(-1)))/(2(2))#

#rArr x=(1pmsqrt(3))/(2)#

#rArr (1+sqrt3)/2, and (1-sqrt3)/2#

You could change these to decimals or leave it in surd form, however, I will leave in surd form.