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

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Answer 1 · 2018-06-06T13:07:36

#rarrsqrt(x-3)+sqrt(x+5)=4#

#rarr(sqrt(x-3))^2=(4-sqrt(x+5))^2#

#rarrcancel(x)-3=16-8sqrt(x+5)cancel(+x)+5#

#rarr8sqrt(x+5)=24#

#rarrsqrt(x+5)=3#

#rarrx+5=9#

#rarrx=4#