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x^2 - 3x -10 > 0 ? Seems pretty easy, but I need someone [#permalink]
09 Sep 2008, 18:59

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

(x-5)(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).

Re: How do you solve this inequality algebraically? [#permalink]
09 Sep 2008, 21:17

also not sure about your rule but agree with the solution.

since (x-5)(x+2) > 0, both terms should be either +ve or -ve. Then only (x-5)(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. _________________

Re: How do you solve this inequality algebraically? [#permalink]
10 Sep 2008, 06:48

bigfernhead wrote:

Can someone prove my Step #3 Wrong? I've tried about 10 different samples, and all of them worked...

(x-5)(x+2) > 0

is possible when both (x-5) and (x-2) are +ve or both negative.

two solutions. (1) x-5>0 and x+2>0 ----> x>5 and x>-2 ----> x>5 (this includes x>-2 also) (2) x-5<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 _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: How do you solve this inequality algebraically? [#permalink]
10 Sep 2008, 20:05

Suresh, scthakur - Try this example, and tell me what solutions you get?

X^2+2x-15<0

x2suresh wrote:

bigfernhead wrote:

Can someone prove my Step #3 Wrong? I've tried about 10 different samples, and all of them worked...

(x-5)(x+2) > 0

is possible when both (x-5) and (x-2) are +ve or both negative.

two solutions. (1) x-5>0 and x+2>0 ----> x>5 and x>-2 ----> x>5 (this includes x>-2 also) (2) x-5<0 and x+2<0 ---> x<5 and x<-2 ---> x<-2(this inclludes x<5 also)

Re: How do you solve this inequality algebraically? [#permalink]
10 Sep 2008, 21:53

can we apply the principle presented in the Step-3 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 ? _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: How do you solve this inequality algebraically? [#permalink]
10 Sep 2008, 22:00

amitdgr wrote:

can we apply the principle presented in the Step-3 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)(x-3) < 0 then either (x+5) < 0 and (x-3) > 0 or (x+5) > 0 and (x-3) < 0

If (x+5) < 0 and (x-3) > 0 then, x < -5 and x > 3 which is illogical. Hence, disregard.

However, if (x+5) > 0 and (x-3) < 0 then x > -5 and x < 3 and hence this will be the solution.

Re: How do you solve this inequality algebraically? [#permalink]
10 Sep 2008, 22:21

scthakur wrote:

amitdgr wrote:

can we apply the principle presented in the Step-3 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)(x-3) < 0 then either (x+5) < 0 and (x-3) > 0 or (x+5) > 0 and (x-3) < 0

If (x+5) < 0 and (x-3) > 0 then, x < -5 and x > 3 which is illogical. Hence, disregard.

However, if (x+5) > 0 and (x-3) < 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 ? _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: How do you solve this inequality algebraically? [#permalink]
11 Sep 2008, 12:02

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.

Re: How do you solve this inequality algebraically? [#permalink]
11 Sep 2008, 19:50

You're right, it doesn't work. Thanks for proving me wrong.

amitdgr wrote:

can we apply the principle presented in the Step-3 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 ?

Re: How do you solve this inequality algebraically? [#permalink]
11 Sep 2008, 20:40

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 _________________

"You have to find it. No one else can find it for you." - Bjorn Borg