How do you solve #abs(4x-3) = 5sqrt(x+4)#?

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Answer 1 · 2016-07-28T11:40:12

#x = 1/32 (49 pm 5 sqrt[329])#

#abs(4x-3) = 5sqrt(x+4)->(4x-3)^2=25abs(x+4)#

or

#16x^2-24x+9=16x^2-25x+x+4+5 = 25abs(x+4)#

or

#(16x^2-25x+5)/(x+4)+1=pm25#

we have two outcomes

#16x^2-25x+5=24(x+4)# resulting in

#x = 1/32 (49 pm 5 sqrt[329])#

and

#16x^2-25x+5=-26(x+4)# with

#x = 1/32 (-1 pm i sqrt[8511])#

the feasible solutions being

#x = 1/32 (49 pm 5 sqrt[329])#