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Maths question: 4((x-2)^4) + 1 = 0?

+2 votes
Solve for x.!!!!!!!!!!!!!!!!!!!!!!!
asked Jan 18, 2013 in ALGEBRA 1 by homeworkhelp Scholar

1 Answer

+3 votes

4((x - 2)^4) +1=0

Subtract 1 from each side.

4((x - 2)^4) + 1 -1=0 -1

.4((x - 2)^4) = -1

Divide each side by 4

4((x - 2)^4)/4 = -1/4

(x - 2)^4) = -1/4

(x - 2)^2*(x - 2)^2= -1/4

(x^2 +4 -4x)*(x^2 +4 -4x)=- 1/4

(x^2 +4 -4x)= -1/4 ,            (x^2 +4 -4x)=- 1/4

(x^2 +4 -4x)= -1/4

Multiply each side by

.(x^2 +4 -4x)*4= (-1/4)*4

.(4x^2 +16 -16x)=-1

Add 1 to each side

.4x^2 +16 -16x+1=-1+1

.4x^2 -16x+17 =0

comparing  above equation ax^2 +bx +c=0

a =4 ,  b =-16 , c =17

    ±

=(-b ± b^2 -4ac)/2a

=(16 ± √16^2 - 4*4*17)/2*4

=(16± √256 -272)/8

 =(16± √-16)/8

=(16±4i)/8           ( i^2=-1)

=(16+4i)/8 ,(16-4i)/8

similarlly

  (x^2 +4 -4x)=- 1/4⇒(16+4i)/8 ,(16-4i)/8

the solution is =(16+4i)/8 ,(16-4i)/8 ,(16+4i)/8 ,(16-4i)/8

 

answered Jan 18, 2013 by krish Pupil

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