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

Answer 1

Kennedy Adams

#color(green)(x=27#

#color(blue)(sqrt(2x+10)-6=2#

Add #6# both sides:

#rarrcolor(blue)(sqrt(2x+10)=2+6#

#rarrcolor(blue)(sqrt(2x+10)=8#

Square both sides to get rid of the radical sign:

#rarrcolor(blue)((sqrt(2x+10))^2=8^2#

#rarrcolor(blue)(2x+10=64#

Subtract #10# both sides:

#rarrcolor(blue)(2x+10-10=64-10#

#rarrcolor(blue)2x=54#

Divide both sides by #2#:

#rarrcolor(blue)((2x)/2=54/2#

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

#rArrcolor(green)(x=27#

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Answer 2

Ethan Adkins

To solve the equation sqrt(2x + 10) - 6 = 2, you can follow these steps:

Add 6 to both sides of the equation: sqrt(2x + 10) = 8.
Square both sides of the equation to eliminate the square root: (sqrt(2x + 10))^2 = 8^2.
Simplify the equation: 2x + 10 = 64.
Subtract 10 from both sides of the equation: 2x = 54.
Divide both sides of the equation by 2: x = 27.

Therefore, the solution to the equation sqrt(2x + 10) - 6 = 2 is x = 27.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.