How do you solve $\sqrt{2 x + 10} - 6 = 2$ $sqrt(2x + 10 )- 6 = 2$ ?

Question

9961 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-14T07:21:10

$x = 27$ $color(green)(x=27$

$\sqrt{2 x + 10} - 6 = 2$ $color(blue)(sqrt(2x+10)-6=2$

Add $6$ $6$ both sides:

$\to \sqrt{2 x + 10} = 2 + 6$ $rarrcolor(blue)(sqrt(2x+10)=2+6$

$\to \sqrt{2 x + 10} = 8$ $rarrcolor(blue)(sqrt(2x+10)=8$

Square both sides to get rid of the radical sign:

$\to {(\sqrt{2 x + 10})}^{2} = 8^{2}$ $rarrcolor(blue)((sqrt(2x+10))^2=8^2$

$\to 2 x + 10 = 64$ $rarrcolor(blue)(2x+10=64$

Subtract $10$ $10$ both sides:

$\to 2 x + 10 - 10 = 64 - 10$ $rarrcolor(blue)(2x+10-10=64-10$

$\to 2 x = 54$ $rarrcolor(blue)2x=54$

Divide both sides by $2$ $2$ :

$\to \frac{2 x}{2} = \frac{54}{2}$ $rarrcolor(blue)((2x)/2=54/2$

$rarrcolor(blue)((cancel2x)/cancel2=27$

$rArrcolor(green)(x=27$

How do you solve sqrt(2x + 10 )- 6 = 2√2x+10−6=2sqrt(2x + 10 )- 6 = 2?