what is the answer in 5squar root of x-4=4-x?

Answer 1

I guess you're saying 5squared root of(x-4)=4-x

5 squared root of (x-4)=4-x
squared root of (x-4)=(4-x)/5 <<<== divide both sides by 5
x-4 = (4-x)^2 / 25 <<<== square both side to get rid of the squared root
25(x-4)= (4-x)^2 <<== multiply both side by 25
25x-100=16-8x+x^2
x^2-25x-8x+100+16=0 <<== move everything to a side to solve the equation
x^2-33x+116=0
(x-4)(x-29) = 0
x=4, x = 29

I hope this would help ^^

Answer 2

If you mean the 5-root then:

5rt(x - 4) = 4 - x
x - 4 = (4 - x)^5 (take 5 power on both sides)
x - 4 + (x - 4)^5 = 0 (bring to the other side)

x = 4 is an answer so eliminate it (divide by (x-4))

1 + (x-4)^4 = 0
(x-4)^4 = -1 this has no real solutions but is has 4 complex solutions.

Answer 3

10

Answer 4

if root belongs only x then
take square
x^2=8-x

x^2+x-8=0
FIND THE ROOTS OF THIS EQUATION


if root belongs to (x-4)
then
take square
x-4=(4-x)^2

x^2-9x+20=0
FIND THE ROOTS OF THIS EQUATION

Answer 5

Jheeeez, darlene: put in some brackets so we can unravel what the .... you mean!

P.S. iyiogrenc... has made two different interpretations but ignored the 5! That CAN't be O.K. surely?!

But in any case: YOU really must TAKE FAR MORE CARE in asking your questions. It's simply NOT FAIR to those trying to help you.

Good night!

Answer 6

0 = 0

?

Answer 7

1.319507911