sqrt(x-3)=sqrt(4(x+4))-1√x−3=√4(x+4)−1
Since we cannot get rid of the 11, let's square and see what we get.
(sqrt(x-3))^2=(sqrt(4(x+4))-1)^2(√x−3)2=(√4(x+4)−1)2
x-3=(sqrt(4(x+4)))^2-2*sqrt(4(x+4))*1+(1)^2x−3=(√4(x+4))2−2⋅√4(x+4)⋅1+(1)2
x-3=4(x+4)-2sqrt(4(x+4))+1x−3=4(x+4)−2√4(x+4)+1
x-3=4x+17-2sqrt(4(x+4))x−3=4x+17−2√4(x+4)
2sqrt(4(x+4))=3x+202√4(x+4)=3x+20
Now, let's square again.
(2sqrt(4(x+4)))^2=(3x+20)^2(2√4(x+4))2=(3x+20)2
4*4(x+4)=(3x)^2+2*(3x)*20+20^24⋅4(x+4)=(3x)2+2⋅(3x)⋅20+202
16x+64=9x^2+120x+40016x+64=9x2+120x+400
9x^2+104x+336=09x2+104x+336=0
Before we use the quadratic formula, let's check the discriminant to see if there are real solutions. If b^2-4acb2−4ac is less than 00, then there are no real solutions; if 00, there is only 11 solution; if greater than 00, there are 22 solutions. You can analyze the quadratic formula to see why this is the case.
(104)^2-4*9*336(104)2−4⋅9⋅336
=-1280=−1280
Thus, there are no real solutions.
