How do you solve #4 sqrt (x) = 8 + 2 sqrt (x)#?

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Answer 1 · 2015-09-13T08:42:42

#color(green)(x = 16#

We are given that #4 sqrt (x) = 8 + 2 sqrt (x)#

Transposing #2sqrtx# to the left hand side, we get:

#4 sqrt (x) - 2 sqrt (x) = 8 #

#(4-2)sqrtx = 8 #

#2sqrtx = 8 #

Divising both sides by 2, we get:

#(cancel(2)sqrtx)/cancel(2) = 8/2 #

#sqrtx = 4#

Squaring both sides we get:

#(sqrtx)^2 = 4^2#

In exponents, #color(blue)(sqrta*sqrta = a#

Hence we get #color(green)(x = 16#