How do you solve using the quadratic formula 2x^2 - 2x = 12x2−2x=1?

2 Answers
Martin C.
Apr 29, 2018

x_1=(1+sqrt(3))/2 or x_2=(1-sqrt(3))/2x1=1+√32orx2=1−√32

Explanation:

2x^2-2x=1|-12x2−2x=1∣−1
2x^2-2x-1=0|:22x2−2x−1=0∣:2
x^2-x-1/2=0x2−x−12=0
x_(1,2)=-(p/2)+-sqrt((p/2)^2-q)x1,2=−(p2)±√(p2)2−q
x_(1,2)=-(-1/2)+-sqrt((-1/2)^2-(-1/2))x1,2=−(−12)±√(−12)2−(−12)
x_(1,2)=1/2+-sqrt(1/4+1/2)=1/2+-sqrt(3)/2x1,2=12±√14+12=12±√32

Brandon
Apr 29, 2018

x=(1pmsqrt3)/2x=1±√32

Explanation:

Minus 11:

2x^2-2x-1=02x2−2x−1=0

Use the formula:

rArr x=-bpm(sqrt(b^2-4ac))/(2a)⇒x=−b±√b2−4ac2a

Use the quadratic to find the values:

2x^2-2x-12x2−2x−1

a=2a=2

b=-2b=−2

c=-1c=−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.

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