How do you solve \sqrt { x - 3} = \sqrt [ 4(x + 4)] - 1?

1 Answer
Oct 15, 2017

No real solutions.

Explanation:

sqrt(x-3)=sqrt(4(x+4))-1

Since we cannot get rid of the 1, let's square and see what we get.

(sqrt(x-3))^2=(sqrt(4(x+4))-1)^2

x-3=(sqrt(4(x+4)))^2-2*sqrt(4(x+4))*1+(1)^2

x-3=4(x+4)-2sqrt(4(x+4))+1

x-3=4x+17-2sqrt(4(x+4))

2sqrt(4(x+4))=3x+20

Now, let's square again.

(2sqrt(4(x+4)))^2=(3x+20)^2

4*4(x+4)=(3x)^2+2*(3x)*20+20^2

16x+64=9x^2+120x+400

9x^2+104x+336=0

Before we use the quadratic formula, let's check the discriminant to see if there are real solutions. If b^2-4ac is less than 0, then there are no real solutions; if 0, there is only 1 solution; if greater than 0, there are 2 solutions. You can analyze the quadratic formula to see why this is the case.

(104)^2-4*9*336

=-1280

Thus, there are no real solutions.