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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
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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 X< 2
Step 4. Confirm Solution
x>5, and x<2
Is this correct for all cases (especially for step 3, which I am not confident about).
Thanks



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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.




Re: How do you solve this inequality algebraically?
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