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x^2  3x 10 > 0 ? Seems pretty easy, but I need someone [#permalink]
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09 Sep 2008, 19:59
x^2  3x 10 > 0 ? Seems pretty easy, but I need someone to confirm if my logic is correct. This is what I learned way back in the HS days. Step 1. Factor (x5)(x+2) > 0 Step 2. Solve X > 5 and X > 2 Step 3. If a solution has a negative sign (in this example it would be x > 2 ), then you flip the sign, and doing so would will you the correct solution. X > 2  flip it and it becomes X5, and x<2 Is this correct for all cases (especially for step 3, which I am not confident about). Thanks == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: How do you solve this inequality algebraically? [#permalink]
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09 Sep 2008, 21:43
I am not sure about the principle in step 3.
However, I will solve it as below.
(x5)(x+2) > 0
This means (x5) and (x+2) both should be greater than 0 or both should be less than zero.
Now, if both are greater than zero then x > 5 and x > 2. This means, x should be greater than 5.
Similarly, if both are smaller than 0 then x < 5 and x < 2. This means, x should be smaller than 2.
Hence, the solution will be x > 5 or x < 2.



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Re: How do you solve this inequality algebraically? [#permalink]
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09 Sep 2008, 22:17
also not sure about your rule but agree with the solution. since (x5)(x+2) > 0, both terms should be either +ve or ve. Then only (x5)(x+2) becomes greater than 0. to achieve bother terms either only +ve or only ve, x must be either greater than 5 or smaller than 2.
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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 00:11
guys, can someone tell me why it is (x5) * (x+2) and not (x+5) * (x2)
many thanks



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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 07:18
Can someone prove my Step #3 Wrong? I've tried about 10 different samples, and all of them worked...



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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 07:48
bigfernhead wrote: Can someone prove my Step #3 Wrong? I've tried about 10 different samples, and all of them worked... (x5)(x+2) > 0 is possible when both (x5) and (x2) are +ve or both negative. two solutions. (1) x5>0 and x+2>0 > x>5 and x>2 > x>5 (this includes x>2 also) (2) x5<0 and x+2<0 > x<5 and x<2 > x<2(this inclludes x<5 also) so final solution is : x<2 or x>5
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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 08:57
bigfernhead wrote: Can someone prove my Step #3 Wrong? I've tried about 10 different samples, and all of them worked... I never looked that way but it seems your rule in step 3 works.
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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 21:05
Suresh, scthakur  Try this example, and tell me what solutions you get? X^2+2x15<0 x2suresh wrote: bigfernhead wrote: Can someone prove my Step #3 Wrong? I've tried about 10 different samples, and all of them worked... (x5)(x+2) > 0 is possible when both (x5) and (x2) are +ve or both negative. two solutions. (1) x5>0 and x+2>0 > x>5 and x>2 > x>5 (this includes x>2 also) (2) x5<0 and x+2<0 > x<5 and x<2 > x<2(this inclludes x<5 also) so final solution is : x<2 or x>5



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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 22:12
bigfernhead wrote: Try this example: X^2+2x15<0 x^2+2x15<0 (x + 5 ) (x  3) < 0 5 < x < 3.
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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 22:53
can we apply the principle presented in the Step3 above when both the solutions are negative eg. x^2+8x+15>0 (x+5)(x+3) > 0 1) when both (x+5) and (x+3) are +ve, we have x > 5 and x > 3 , this essentially means x > 3 2) when both (x+5) and (x+3) are ve, we have x < 5 and x < 3 , this essentially means x < 5 so 3 < x <5 If we try and apply step 3 on solving the equation we get (x+5)(x+3) > 0 x> 5 and x>3 so reversing the signs we get x<5 and x<3 i guess it is not working or did I miss something ?
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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 23:00
amitdgr wrote: can we apply the principle presented in the Step3 above when both the solutions are negative eg. x^2+8x+15>0 (x+5)(x+3) > 0 1) when both (x+5) and (x+3) are +ve, we have x > 5 and x > 3 , this essentially means x > 3 2) when both (x+5) and (x+3) are ve, we have x < 5 and x < 3 , this essentially means x < 5 so 3 < x <5 If we try and apply step 3 on solving the equation we get (x+5)(x+3) > 0 x> 5 and x>3 so reversing the signs we get x<5 and x<3 i guess it is not working or did I miss something ? I would prefer the basic principle. If the multiplication of two algebraic expressions is negative, one of them will be negative and another one will be positive. Thus, if (x+5)(x3) < 0 then either (x+5) < 0 and (x3) > 0 or (x+5) > 0 and (x3) < 0 If (x+5) < 0 and (x3) > 0 then, x < 5 and x > 3 which is illogical. Hence, disregard. However, if (x+5) > 0 and (x3) < 0 then x > 5 and x < 3 and hence this will be the solution.



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Re: How do you solve this inequality algebraically? [#permalink]
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10 Sep 2008, 23:21
scthakur wrote: amitdgr wrote: can we apply the principle presented in the Step3 above when both the solutions are negative eg. x^2+8x+15>0 (x+5)(x+3) > 0 1) when both (x+5) and (x+3) are +ve, we have x > 5 and x > 3 , this essentially means x > 3 2) when both (x+5) and (x+3) are ve, we have x < 5 and x < 3 , this essentially means x < 5 so 3 < x <5 If we try and apply step 3 on solving the equation we get (x+5)(x+3) > 0 x> 5 and x>3 so reversing the signs we get x<5 and x<3 i guess it is not working or did I miss something ? I would prefer the basic principle. If the multiplication of two algebraic expressions is negative, one of them will be negative and another one will be positive. Thus, if (x+5)(x3) < 0 then either (x+5) < 0 and (x3) > 0 or (x+5) > 0 and (x3) < 0 If (x+5) < 0 and (x3) > 0 then, x < 5 and x > 3 which is illogical. Hence, disregard. However, if (x+5) > 0 and (x3) < 0 then x > 5 and x < 3 and hence this will be the solution. Thanks scthakur , how would you solve for (x+5)(x+3) > 0 ?
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Re: How do you solve this inequality algebraically? [#permalink]
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11 Sep 2008, 13:02
I think this problem is easy if you just graph it.
The graph of this equation is a parabola with x intercepts 5 and 2. Additionally, since the leading term in the equation is positive the parabola opens up, like a U.
Thus, the value of this equation must be greater than 0 when x < 2 or x > 5.
Please tell me if this makes sense.



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Re: How do you solve this inequality algebraically? [#permalink]
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11 Sep 2008, 20:50
You're right, it doesn't work. Thanks for proving me wrong. amitdgr wrote: can we apply the principle presented in the Step3 above when both the solutions are negative eg. x^2+8x+15>0 (x+5)(x+3) > 0 1) when both (x+5) and (x+3) are +ve, we have x > 5 and x > 3 , this essentially means x > 3 2) when both (x+5) and (x+3) are ve, we have x < 5 and x < 3 , this essentially means x < 5 so 3 < x <5 If we try and apply step 3 on solving the equation we get (x+5)(x+3) > 0 x> 5 and x>3 so reversing the signs we get x<5 and x<3 i guess it is not working or did I miss something ?



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Re: How do you solve this inequality algebraically? [#permalink]
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11 Sep 2008, 21:40
For quadratic inequalities 1) ax² + bx + c > 0 , where a > 0 and say p and q are the roots ( q having smaller value and p the larger value) then the inequality will hold good for values of x where x>p and x<q, in other words ; when the inequality sign is greater than, the value of x does not lie between the two roots p and q 2) ax² + bx + c < 0 , where a > 0 and say p and q are the roots ( q having smaller value and p the larger value) then the inequality will hold good for values of x where q < x < p, in other words ; when the inequality sign is lesser than , the value of x lies between the two roots p and q
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Re: How do you solve this inequality algebraically? [#permalink]
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12 Sep 2008, 05:28
amitdgr wrote: Thanks scthakur , how would you solve for (x+5)(x+3) > 0 ? Now I see the pattern here.....however, your step 3 will not be true....as the solution here will be either x > 3 or x < 5. However, your generic rules as defined in terms of ax^2 + bx + c are perfectly applicable. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.




Re: How do you solve this inequality algebraically?
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