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How do you solve #|x+2|= 4#?

Answer 1

#x=-6color(white)("XX")orcolor(white)("XX")x=2#

Consider the two possibilities: #{: (x+2 < 0,color(white)("XX")andcolor(white)("XX"),x+2 > 0), (rarr abs(x+2) = -x-2,,rarrabs(x+2)=x+2), ("so "abs(x+2)=4,,"so "abs(x+2)=4), (rarr -x-2 =4,,rarr x+2=4), (rarr -x=6,,rarrx==2), (rarr x=-6,,) :}#

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Answer 2

Nathan Axel

#x=2" or " x=-6#

Equations with an#color(blue)" absolute value"# normally have 2 solutions.

These are found by solving #x+2=color(red)(+-)4#

#color(blue)"Solution 1"#

#"solve "x+2=4rArrx=4-2=2#

#color(blue)"Solution 2"#

#"solve " x+2=-4rArrx=-4-2=-6#

#color(blue)"As a check"#

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

#x=2to|2+2|=|4|=4=" right side"#

#x=-6to|-6+2|=|-4|=4=" right side"#

#rArrx=2" or "x=-6" are the solutions"#

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Answer 3

Jace Anderson

To solve the equation |x + 2| = 4, you need to consider two cases: when the expression inside the absolute value is positive and when it is negative.

Case 1: x + 2 = 4 Solve for x: x = 4 - 2 = 2

Case 2: -(x + 2) = 4 Solve for x: x + 2 = -4 x = -4 - 2 x = -6

The solutions are x = 2 and x = -6.

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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.