How do you solve #4= sqrt x - 3root4x#?

Question

1345 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-14T13:48:28

#x=4^4=256#

#4= sqrtx - 3root4x#

Set #y=root4x#, #sqrtx# became #y^2#. Hence,

#y^2-3y=4#

#y^2-3y-4=0#

#(y-4)*(y+1)=0#

So #y_1=-1# and #y_2=4#

But #x=(-1)^4=1# doesn't provide equation. Thus,

#x=4^4=256# is solution of it.

Answer 2 · 2018-03-14T13:49:04

#" The Soln. Set "sub RR" is "{256}#.

We will solve the eqn. in #RR#.

Letting, #root(4)x=t, x=t^4 :. sqrtx=t^2#.

Substituting in the given eqn., we get,

#4=t^2-3t, or, t^2-3t-4=0#.

#:. ul(t^2-4t)+ul(t-4)=0#.

#:. t(t-4)+1(t-4)=0#.

#:. (t-4)(t+1)=0#.

#:. t=4, or, t=-1#.

#:. x=t^4=4^4=256, or, x=(-1)^4=1#.

But, we observe that, #x=1# does not satisfy the original eqn.

#:." The Soln. Set "sub RR" is "{256}#.