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C3 inequality with modulus

image.jpgStruggling with the inequality part of this question, set 2(x+3) < +/- x but I get two values x < -6 and x < -2 which doesn't seem right or worth 5 marks
the answer is actually -6<x<-2 . My best tip for doing inequality questions (with or without a modulus sign) is to draw a graph. when you draw the graph, it becomes obvious that |x| is only greater than |2(x+3)| when x is between -6 and -2
when you have modulus on both sides, square both sides

|x|>|2(x+3)|

x^2 - (2(x+3))^2 > 0

-3x^2 - 24x - 36 > 0

x^2 + 8x + 12 < 0

(x+2)(x+6) < 0

x = -2 and x = -6



therefore -6<x<-2
(edited 8 years ago)
Another way to do it is, firstly remove the greater sign and equate the equations and find the critical values. This will give you good idea on from where to where you need to draw your graph

easiest way is to just square both sides
|x|=|2(x+3)|x^2 = (2(x+3))^2-3x^2 - 24x - 36 = 0x^2 + 8x + 12 = 0(x+2)(x+6) = 0

x=-2 and x=-6

so graphing from like x=-8 to x=0 would show you both intersections and a little extra



You can see |x| is above |2(x+3)| between -2 and -6
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Student20
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Original post by Bealzibub
Another way to do it is, firstly remove the greater sign and equate the equations and find the critical values. This will give you good idea on from where to where you need to draw your graph

easiest way is to just square both sides
|x|=|2(x+3)|x^2 = (2(x+3))^2-3x^2 - 24x - 36 = 0x^2 + 8x + 12 = 0(x+2)(x+6) = 0

x=-2 and x=-6

so graphing from like x=-8 to x=0 would show you both intersections and a little extra



You can see |x| is above |2(x+3)| between -2 and -6



Ah I see now, thank you

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