x^2 > 2xmarie wrote:as part of a ds problem I had to resolve:
x^2 > 2x i.e., {x squared is greater than 2x}
is the answer: x>2 and x<0 or x>2 and x<-2??
thxs,
please show all steps!
x^2 - 2x > 0
x(x - 2) > 0
0 > x > 2
inequalities
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JeffB
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x^2 - 2x > 0
x(x - 2) > 0
0 > x > 2
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Stuart@KaplanGMAT
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As JeffB notes, we can simplify to:marie wrote:as part of a ds problem I had to resolve:
x^2 > 2x i.e., {x squared is greater than 2x}
is the answer: x>2 and x<0 or x>2 and x<-2??
thxs,
please show all steps!
x(x-2)>0
Let's see how we get from there to the final answer.
First, we recognize that to obtain a positive product, either both terms are positive or both are negative. Accordingly, we can set up two pairs of inequalities.
The positive case gives us:
x>0 AND x-2>0
x>0 AND x>2
When we have two inequalities pointing in the same direction, we must satisfy the more limiting inequality. In this case, that means that x>2.
The negative case gives us:
x<0 AND x-2<0
x<0 AND x<2
Again choosing the more limiting inequality, we get x<0.
So, to satisfy the original inequality, either:
x<0 OR x>2.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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Stuart@KaplanGMAT
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Assuming x is negative doesn't mean we replace it with "-x" in the equation; if x is negative and you write "-x", you're actually looking at a double negative, which is positive.marie wrote:Stuart,
Thank you for that. Help me understand where the following line of reasoning goes wrong.
x^2 > 2x
assume x is negative therefore
(-x)(-x) > 2(-x)
divide both sides by -x
-x < 2
multiply by -1
x > -2
Thxs
If we assume x is negative, we leave it as written, we just remember to flip the inequality when we divide by x. So:
x*x>2x
dividing by x:
x<2
which is trivial - all we've shown is that if x is negative, then it's less than 2, which is true of all negative numbers.
I guess the more fundamental issue is why you're assuming that x is negative - how does that help you solve this problem?
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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