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

Question

4128 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-06T13:07:36

$\to \sqrt{x - 3} + \sqrt{x + 5} = 4$ $rarrsqrt(x-3)+sqrt(x+5)=4$

$\to {(\sqrt{x - 3})}^{2} = {(4 - \sqrt{x + 5})}^{2}$ $rarr(sqrt(x-3))^2=(4-sqrt(x+5))^2$

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

$rarr8sqrt(x+5)=24$

$rarrsqrt(x+5)=3$

$rarrx+5=9$

$rarrx=4$

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