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Polynomials Question (1 Viewer)

cutemouse

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Could someone please help me with this question?

The polynomial P(x) is given by P(x)=2x<sup class="moz-txt-sup">3</sup> - 9x<sup class="moz-txt-sup">2</sup> + 12x - k (where k is real). Find the range of values for k for which P(x)=0 has 3 real roots.

Thanks
 

Iruka

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Find the x values of the two stationary points.

Find conditions on k that ensure that the two stationary points have y-values that are opposite in sign.

(Drawing a picture might help.)
 

Timothy.Siu

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Could someone please help me with this question?

The polynomial P(x) is given by P(x)=2x<sup class="moz-txt-sup">3</sup> - 9x<sup class="moz-txt-sup">2</sup> + 12x - k (where k is real). Find the range of values for k for which P(x)=0 has 3 real roots.

Thanks
P'(x )=6x2-18x+12
=6(x2-3x+2)
=6(x-2)(x-1)
=0 when x=2 or 1

(just to check)
P''(x )=6(2x-3)
when x=2 P''(x)=6 , minimum turning point.
when x=1 P''(x)=-6, max. turning point.

ok.
x=1 P(x)=5-k
x=2 P(x)=4-k

i'm not sure if this is the best way to do it.
5-k>0
and 4-k<0
(since we want 3 real roots)
5>k and 4<k
therefore 4<k<5
i think.
 

lyounamu

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ye i reckon timmy's approach is right =)

well done
 

cutemouse

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lol, I'm lost... Could anyone else explain it?

I mean, I could do this graphically I guess... ie. by sketching y = 2x<sup class="moz-txt-sup">3</sup> - 9x<sup class="moz-txt-sup">2</sup> + 12x (showing all stat points) against y=k and then by noting where the graphs intesect 3 times... But IMO that's a waste of time and there should be a different/faster way to do it.
 

clintmyster

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P'(x )=6x2-18x+12
=6(x2-3x+2)
=6(x-2)(x-1)
=0 when x=2 or 1

(just to check)
P''(x )=6(2x-3)
when x=2 P''(x)=6 , minimum turning point.
when x=1 P''(x)=-6, max. turning point.

ok.
x=1 P(x)=5-k
x=2 P(x)=4-k

i'm not sure if this is the best way to do it.
5-k>0
and 4-k<0
(since we want 3 real roots)
5>k and 4<k>
therefore 4<k><5
i think.
graphically that makes sense. Can you just use graphs for this question saything that it has to be between those two values as otherwise you will get a double root and another root?

</k></k>

lol, I'm lost... Could anyone else explain it?

I mean, I could do this graphically I guess... ie. by sketching y = 2x3 - 9x2 + 12x (showing all stat points) against y=k and then by noting where the graphs intesect 3 times... But IMO that's a waste of time and there should be a different/faster way to do it.
well tim essentially could sketch it really quickly using what he found for the turning points. if you realise like i did that 3 roots occurs between the values for k where theres a double root and another single root, its quite easy. If you can see it visually in your mind you could do a very very rough sketch and save time.
 
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azureus88

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if it has 3 real roots, then the y-values of the 2 turning points must be opposite in sign. So:

P(1).P(2)<0
(5-k)(4-k)<0

4 < k < 5
 

Iruka

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graphically that makes sense. Can you just use graphs for this question saything that it has to be between those two values as otherwise you will get a double root and another root?

</k></k>



well tim essentially could sketch it really quickly using what he found for the turning points. if you realise like i did that 3 roots occurs between the values for k where theres a double root and another single root, its quite easy. If you can see it visually in your mind you could do a very very rough sketch and save time.
There is a question in the 2001 2u HSC paper (Q6(c)) which uses a similar idea, but with a different cubic.
 

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