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Hi guys, I created this topic to share with you a quick and easy way to solve inequalities(second-degree).

Straightaway here's my tip:

Attachment:

eq.jpg [ 27.79 KiB | Viewed 34316 times ]

This works with \(\geq{}\) and \(\leq{}\) as well, you just add the = sign, that's it.

The METHOD I) replace the\(>,<\) sign with a more friendly \(=\) II) find \(x1\) , \(x2\) as you would do normally III) use my trick to define the interval.

If the sign of \(a\) and the operator are "the same" (>,+) or (<,-) we take the ESTERNAL values. Otherwise the INTERNAL values.

I find this method very easy to remember and to use: I'm sure it will save you time. This is it, but for those of you interested in what goes on behind the scene I have a mathematical explanation.

\(ax^2+bx+c\) is a parabola.

I took two parabolas one with a positive \(a\), one with a negative \(a\). \(a\) in the formulas defines 2 characteristics of the parabola: 1) its "slope" ( greater \(a\) greater the slope) 2) where the parabola looks at ( \(a\) +ve the parabola looks up, \(a\) -ve the parabola looks down).

Attachment:

a.jpg [ 51.24 KiB | Viewed 34372 times ]

With this little theory, and looking to the graphs we can conclude that: if \(a\) is +ve and we want the values >0 we have to take the esternal values, if we want the values <0 the internal values. This same principle can be applyied to a negative \(a\)

Some examples: \(f(x)=x^2+2x-3>0\) Step 1) replace > with = : \(x^2+2x-3 = 0\) Step 2) find x1 and x2: \(x1=1, x2=-3\) Step 3)use the "tip": \(a\) is +ve and the operator is > solution \(x>1\) and \(x<-3\) \(f(x)=x^2+2x-3<0\) same function diff operator all the steps are the same Step 3)use the "tip": \(a\) is +ve and the operator is < solution \(-3<x<1\)

Hi Zarrolou , Thanks for your tip. Can you please elaborate it with more examples? I have difficulty in understanding it.

Thanks.

Sure! Here are all possible cases: 1)\(x^2-3x+2>0\) 2)\(x^2-3x+2<0\)

The roots of \(x^2-3x+2=0\) are x=1 and x=2 In 1 the sign of x^2 is + and the operator is > => they are "the same" so we take the external values \(x<1\) and \(x>2\) In 2 the signs are not the same => we take the internal values \(1<x<2\)

in these the sign of x^2 is - 3)\(-2x^2+3x+2>0\) 4)\(-2x^2+3x+2<0\)

The roots of \(-2x^2+3x+2 =0\) are \(x=2\) and \(x=-\frac{1}{2}\). In 3 the sign of x^2 is - and the operator is > => the are not the same so we take the internal values \(-\frac{1}{2}<x<2\) In 4 they are the same (-,<) external values \(x<-\frac{1}{2}\) and \(x>2\)

Hope it's clear now. Let me know _________________

It is beyond a doubt that all our knowledge that begins with experience.

For #1, the roots are both positive (1, 2) and thus are greater than (>) zero. The inequality itself is greater than zero, therefore we take the external values (x<1 & x>2)

For #2, the roots are both positive (1, 2) and thus are greater than (>) zero. However the inequality here is less than zero, therefore we take the values between the two toots (1<x<2)

But what happens if one root is positive and one root is negative?

For #1, the roots are both positive (1, 2) and thus are greater than (>) zero. The inequality itself is greater than zero, therefore we take the external values (x<1 & x>2)

For #2, the roots are both positive (1, 2) and thus are greater than (>) zero. However the inequality here is less than zero, therefore we take the values between the two toots (1<x<2)

But what happens if one root is positive and one root is negative?

Thanks!

. Let me know

Not quite.

I am refering to the sign of a in +-a\(x^2+bx+c\), not to the sign of the roots. You have to look at the original equation.

If that sign and the operator are the "same"(+ with >, - with <), take the exsternal values; if the are different, take the internal values. _________________

It is beyond a doubt that all our knowledge that begins with experience.

Hi guys, I created this topic to share with you a quick and easy way to solve inequalities(second-degree).

Straightaway here's my tip:

Attachment:

eq.jpg

This works with \(\geq{}\) and \(\leq{}\) as well, you just add the = sign, that's it.

The METHOD I) replace the\(>,<\) sign with a more friendly \(=\) II) find \(x1\) , \(x2\) as you would do normally III) use my trick to define the interval.

If the sign of \(a\) and the operator are "the same" (>,+) or (<,-) we take the ESTERNAL values. Otherwise the INTERNAL values.

I find this method very easy to remember and to use: I'm sure it will save you time. This is it, but for those of you interested in what goes on behind the scene I have a mathematical explanation.

\(ax^2+bx+c\) is a parabola.

I took two parabolas one with a positive \(a\), one with a negative \(a\). \(a\) in the formulas defines 2 characteristics of the parabola: 1) its "slope" ( greater \(a\) greater the slope) 2) where the parabola looks at ( \(a\) +ve the parabola looks up, \(a\) -ve the parabola looks down).

Attachment:

a.jpg

With this little theory, and looking to the graphs we can conclude that: if \(a\) is +ve and we want the values >0 we have to take the esternal values, if we want the values <0 the internal values. This same principle can be applyied to a negative \(a\)

Some examples: \(f(x)=x^2+2x-3>0\) Step 1) replace > with = : \(x^2+2x-3 = 0\) Step 2) find x1 and x2: \(x1=1, x2=-3\) Step 3)use the "tip": \(a\) is +ve and the operator is > solution \(x>1\) and \(x<-3\) \(f(x)=x^2+2x-3<0\) same function diff operator all the steps are the same Step 3)use the "tip": \(a\) is +ve and the operator is < solution \(-3<x<1\)

If you like the tip and you're gonna use it, give it a KUDOS!

Hope you guys find it useful, regards

Can you elaborate more on the external and internal values ? That should clear the concept completely for me. Is it possible for you to draw a shaded area on the graph that explains what you mean by the external and internal values as defined by x1 and x2 ? _________________

Yogi Bhajan: If you want to learn a thing, read that; if you want to know a thing, write that; if you want to master a thing, teach that. This message transmitted on 100% recycled electrons.

Can you elaborate more on the external and internal values ? That should clear the concept completely for me. Is it possible for you to draw a shaded area on the graph that explains what you mean by the external and internal values as defined by x1 and x2 ?

For any quadratic equation \(ax^2\) + bx + c = 0, we almost always get two roots.(say x1 and x2) The internal value means the values of x that falls between x1 and x2. And the external value mean the values of x that are greater than x2 ox less than x1 (Assuming that x1 < x2).

For example, the quadratic equation \(x^2\) - 8x + 12 = 0 will give you the roots as 2 and 6.

we get 3 intervals of x. i) x < 2 ii) 2 < x < 6 iii) x > 6 of which values that are beyond the boundaries of roots are external values i.e. x < 2 and x > 6 and that are within the boundaries of roots are internal values 2 < x < 6. Now to determine which interval to select, use the method presented by Zarrolou.

Can you elaborate more on the external and internal values ? That should clear the concept completely for me. Is it possible for you to draw a shaded area on the graph that explains what you mean by the external and internal values as defined by x1 and x2 ?

For x_1=-10 and x_2=-1 External values: ------------------(-10)-------------(-1)------------------ Internal values: ------------(-10)------------------(-1)------------------

External values=" values greater then the greatest root, and smaller than the smallest root". Internal values="values in between the two roots".

In the image below there are two graphical examples. The first one represents the solution for \(x^2-8x+15>0\) The second one for \(-x^2+8x-15>0\)

Hope everything is clear, let me know. Thanks

Attachments

Imm.JPG [ 53.54 KiB | Viewed 31631 times ]

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Can you elaborate more on the external and internal values ? That should clear the concept completely for me. Is it possible for you to draw a shaded area on the graph that explains what you mean by the external and internal values as defined by x1 and x2 ?

For x_1=-10 and x_2=-1 External values: ------------------(-10)-------------(-1)------------------ Internal values: ------------(-10)------------------(-1)------------------

External values=" values greater then the greatest root, and smaller than the smallest root". Internal values="values in between the two roots".

In the image below there are two graphical examples. The first one represents the solution for \(x^2-8x+15>0\) The second one for \(-x^2+8x-15>0\)

Hope everything is clear, let me know. Thanks

Thank you for the explanation. This is more clear to me now. The above is a quadratic equation with one variable. This may seem a very basic question as I have not plotted the quadratic equation myself yet. But can we plot an equation with one variable on a two variable axis ? _________________

Yogi Bhajan: If you want to learn a thing, read that; if you want to know a thing, write that; if you want to master a thing, teach that. This message transmitted on 100% recycled electrons.

Thank you for the explanation. This is more clear to me now. The above is a quadratic equation with one variable. This may seem a very basic question as I have not plotted the quadratic equation myself yet. But can we plot an equation with one variable on a two variable axis ?

You can plot something like

\(y=x^2+5x-15\)=> it's a parabola.

You then can read the graph in this way: 1)where it intersects the x-axis the equation has its root(s). 2)if you are asked \(y=x^2+5x-15>0\), you'll pick the "parts"/range of values for x where the graph is positive or over the y-axis. 3)if you are asked \(y=x^2+5x-15<0\), you'll pick the part where the graph is below the y-axis. 4)if you are asked \(y=x^2+5x-15=0\), you'll pick the single point(s) where the graph intersect the y axis. _________________

It is beyond a doubt that all our knowledge that begins with experience.

(1) x^2 - 4x + 3 < 0 The roots are 3 and 1. Because the sign of \(x^2\) is +, and we have \(<\) ==> INTERNAL values. \(1<x<3\), but we are told that x is an integer, so \(x\) is \(2\). Sufficient

(2) x^2 + 4x +3 > 0 The roots are -1 and -3. Because the sign of \(x^2\) is +, and we have \(>\) ==> EXTERNAL values. \(x>-1\) or \(x<-3\), this is not sufficient.

If x is an integer, is |x| > 1? Or: is x>1 or x<-1 ?

(1) (1 - 2x)(1 + x) < 0 The roots are \(-1\) and \(\frac{1}{2}\). Because if you expand the expression the sign of \(x^2\) is negative, and we have \(<\) ==> External values, so \(x<-1\) or \(x>\frac{1}{2}\). This is not sufficient to answer the question.

(2) (1 - x)(1 + 2x) < 0 The roots are \(1\) and \(-\frac{1}{2}\). Because if you expand the expression the sign of \(x^2\) is negative, and we have \(<\) ==> External values,\(x<-\frac{1}{2}\) or \(x>1\). This is not sufficient to answer the question as well.

Combining the two intervals we obtain the common part, \(x<-1\) or \(x>1\) hence the answer is C (as this directly answers the main question). _________________

It is beyond a doubt that all our knowledge that begins with experience.

This and many more tricks which I discovered here on GMAT Club off late are turning out to be a life saver. GMAT in 10 days and am wondering why wasn't I here earlier

This and many more tricks which I discovered here on GMAT Club off late are turning out to be a life saver. GMAT in 10 days and am wondering why wasn't I here earlier

The Critical points for numerator -1/3, 2/5 Critical point for denominator 0

Total critical points ------------- -1/3 --------------- 0 --------------- 2/5 ---------------

Since the sign of original inequality is positive, the expression will be positive in the rightmost region and in other regions it will be alternatively negative and positive.

That means ---------------- -1/3 +++++++++ 0 ---------------- 2/5 ++++++++++++

Hence Solution of the inequality is -1/3 < x 0 , x > 2/5

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