How do you solve $8 = \sqrt{2 x - 3} + 3$ $8=sqrt( 2x-3)+3$ ?

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Answer 1 · 2016-01-31T08:24:29

$\sqrt{2 x - 3} + 3 = 8$ $sqrt(2x-3)+3=8$

$\to \sqrt{2 x - 3} = 8 - 3$ $rarrsqrt(2x-3)=8-3$

$\to \sqrt{2 x - 3} = 5$ $rarrsqrt(2x-3)=5$

Square both sides:

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

$\to 2 x - 3 = 25$ $rarr2x-3=25$

$\to 2 x = 5 + 3$ $rarr2x=5+3$

$\to 2 x = 8$ $rarr2x=8$

$\to x = \frac{8}{2} = 4$ $rarrx=8/2=4$

How do you solve 8=√2x−3+38=sqrt( 2x-3)+3?