1997 4U question 5c) (1 Viewer)

asianese · Aug 18, 2012

I saw the solution in the blue mansw book a while back but wasn't quite sure.

With v) the answer was b<0. Here is my working (it is a bit long winded but meh)

$For real roots, one stationary point must be below or on the axis. Stationary points occur when $ P'(z)=0 \Rightarrow 4z^3+2bz=0 \\ 2z(2z^2+b)=0\\ z=0 $ or $2z^2=-b \\ $Since b is real, there are no solutions to $2z^2=-b,$ so $z=0 \\ P(0)=d \therefore d\leq0, $ but $ d\neq 0, $ so $ d<0. \\ d=\frac{b^2}{4} <0 $ which has no real solution. So no value of b can give $ P(z) $ real roots.$

(There's obviously a flaw somewhere.)

I think i've assumed b is positive?...I think that's the trick?

Is it because $2z^2=-b$ must have a real solution so b must be <0?

bleakarcher · Aug 18, 2012

asianese said:
I saw the solution in the blue mansw book a while back but wasn't quite sure.

With v) the answer was b<0. Here is my working (it is a bit long winded but meh)

$For real roots, one stationary point must be below or on the axis. Stationary points occur when $ P'(z)=0 \Rightarrow 4z^3+2bz=0 \\ 2z(2z^2+b)=0\\ z=0 $ or $2z^2=-b \\ $Since b is real, there are no solutions to $2z^2=-b,$ so $z=0 \\ P(0)=d \therefore d\leq0, $ but $ d\neq 0, $ so $ d<0. \\ d=\frac{b^2}{4} <0 $ which has no real solution. So no value of b can give $ P(z) $ real roots.$

(There's obviously a flaw somewhere.)

I think i've assumed b is positive? Is there another way to do this?

Well, by the quadratic formula: z^2=[-b+/-sqrt[b^2-4d]]/2=-b/2. Now z^2>=0 if the polynomial is to have real roots i.e. -b/2>=0. So P(z) has real roots if b<0 since d=/=0. Final answer: b<0.

Sindivyn · Aug 18, 2012

asianese said:
I saw the solution in the blue mansw book a while back but wasn't quite sure.

With v) the answer was b<0. Here is my working (it is a bit long winded but meh)

$For real roots, one stationary point must be below or on the axis. Stationary points occur when $ P'(z)=0 \Rightarrow 4z^3+2bz=0 \\ 2z(2z^2+b)=0\\ z=0 $ or $2z^2=-b \\ $Since b is real, there are no solutions to $2z^2=-b,$ so $z=0 \\ P(0)=d \therefore d\leq0, $ but $ d\neq 0, $ so $ d<0. \\ d=\frac{b^2}{4} <0 $ which has no real solution. So no value of b can give $ P(z) $ real roots.$

(There's obviously a flaw somewhere.)

I think i've assumed b is positive?...I think that's the trick?

Is it because $2z^2=-b$ must have a real solution so b must be <0?

For part v, yes.

asianese · Aug 18, 2012

Ah ic thanks both of you. also Bleakarcher, b can't equal 0 because that implies d=0 which is a restriction on the question. But other than that, thanks

bleakarcher · Aug 18, 2012

asianese said:
Ah ic thanks both of you. also Bleakarcher, b can't equal 0 because that implies d=0 which is a restriction on the question. But other than that, thanks

yeah I realised.

1997 4U question 5c) (1 Viewer)

asianese

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bleakarcher

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Sindivyn

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asianese

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bleakarcher

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