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

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Answer 1 · 2018-06-06T13:07:36

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

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

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

$rarr8sqrt(x+5)=24$

$rarrsqrt(x+5)=3$

$rarrx+5=9$

$rarrx=4$

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