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The set of values of a for which the equation x^3 - 3x + a has one negative and two positive roots is ? 

I know that a>0 and i also know that the turning point of the graph is at x=-1 and x = 1 but how to proceed further??

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Has your studies included curve sketching?

 

So sketch the graph of x3 -3x hint can this be factorised?

now a is a 'constant' but we may choose different values of  'a'

What do you think this does to your sketch?

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I know a bit about curve sketching only! So please can u explain more. I know that graph will turn at -1 and then 1.

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4 minutes ago, Shahroze said:

I know that graph will turn at -1 and then 1.

You have said that already.

What you didn't do was answer my question or even attempt it.

Never mind. I didn't say the graph of y = x3 -3x  and I should have.

Right so get the graph of y = x3 -3x + 0    (that is with a = 0)

Now if you add a = a positve value to this then it moves the sketch which way up or down the y axis?

To do your sketch what happens to y if x is very large and negative?

What happens if to y if x is very large and positive?

What happens to y if x = 0?

You have already identified two other points, x = 1 and x = -1

 

 

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Oh yes sorry i didn't understood u yes the graph should move up and down as the points - 1 and 1 are fixing the graph's turning point. If x be very large then y will also be large as it will meet the line of graph will be a slightly straight line.  If a = 0 and x= 0 then y will be zero.  Am i going the right direction please check my sketch!   When i add postive value of a then the graph is moving up!

IMG_20181103_003136674.jpg

Oh.. now i got the logic so do u mean that the extreme values of a should be such that the graph lies between fig1 and fig2 . Oh i am not able to upload the images because of file size but now i understood that we should take those value of a such that graph become such that the turning point touches 1 and -1 and not beyond that as the roots will then become imaginary.  So is the value of a for the above case -2<a<2 ??

 

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Yea that's the idea.

Your sketch slides up and down the y axis (but not left or right) according to teh value of a.

Now look at your graph of y = x3 -3x + a

You are asked for the limits where there is one negative and two positive roots.

First the negative root.

What happens as your sketch slide downwards? (a negative)

Eventually the maximum at x = -1 touches the x axis.

If the modulus of a gets any bigger then that max is below the x axis and there is no root.

Similarly for the minimum, at x = 1

So you need to determine the values of a for these conditions.

If a lies outside these values there are not going to be the roots you are asked for.

 

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