The Student Room Group

Modulus Inequality Question

I've been trying to figure out where I went wrong for ages :frown:
Could someone pls help me
(edited 1 year ago)
You could have checked your answer by subbing a value like x=0 into the original inequality, but your working look ok until the end. Ive have multiplied through by -1 and solved >= 0. Your quadratic is a "n" which is <= 0, so it must be the two tails.

Not sure how you concluded the middle section? Note a sketch at the start might have helped. The lhs is a "v" centered on 2. The right hand side is also a "v" but twice as steep and centered on 3. The twice as steep one will beat the other one for "large" |x|, which gives the answer..
(edited 1 year ago)
Original post by mqb2766
You could have checked your answer by subbing a value like x=0 into the original inequality, but your working look ok until the end. Ive have multiplied through by -1 and solved >= 0. Your quadratic is a "n" which is <= 0, so it must be the two tails.

Not sure how you concluded the middle section?


So I went wrong because I didn't multiply by -1?
Original post by action123
So I went wrong because I didn't multiply by -1?


No your reasoning (whatever it was) was wrong, but the reasoning may have been easier if youd had a "u" quadratic (positive x^2 coefficient).
(edited 1 year ago)
Original post by mqb2766
No your reasoning (whatever it was) was wrong, but the reasoning may have been easier if youd had a "u" quadratic (positive x^2 coefficient).


Sorry but I'm still quite confused because all I did was rearrange the expression after I expanded the brackets
(edited 1 year ago)
Original post by action123
Sorry but I'm still quite confused because all I did was rearrange the expression after I expanded the brackets


Your penultimate line (last but one) is correct. How did you then conclude that the answer was between the two roots, rather than either side of them?
Original post by mqb2766
Your penultimate line (last but one) is correct. How did you then conclude that the answer was between the two roots, rather than either side of them?

I thought that because of the inequality sign '<'
Original post by action123
I thought that because of the inequality sign '<'


If you sketched the quadratic, what would it look like?
Something like this
Original post by action123
Something like this


Correct, so the middle section (between the roots) is > 0. Before and after the roots its < 0.
Ah yes thank you!
Original post by action123
I always thought that if you had a quadratic<0, then you would look at the inside region so would use '≤x≤'
So this part of my reasoning is wrong?

Yes, that only applies with a "u" quadratic so a positive x^2 coefficient. There are several ways to check, some mentioned above, but
* Check, Id have subbed x=0 in. This is outside your answer but clearly 2<6 so 0 is a solution.
* Sketch the functionns (absolute value and / or quadratic). For large |x|, |x| < |2x| so it must be the tails, not the middle region.
* Write down the factors and think where they must be positive/negative to satisfy the relationship
* Id have taken the left over to the right after squaring to get a 0 <= "u" quadratic or multiplied by -1
* ...

Also you dont have to square things up to solve an absolute value inequality, you can simply reason about the sign of the terms and treat it as a few simple linear inequalities.
(edited 1 year ago)

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