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Polynomials Questions (1 Viewer)

MC Squidge

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1a) show that x^n-bx^2+c has a multiple root if (n^n)(c^n-2)=(4b^n)((n-2)^(n-2))
b)Show co4@=8(cos@-cosPI/8)(cos@-cos3PI/8)(cos@-cos5PI/8)(cos@-cos7PI/8)
 

Timothy.Siu

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1a) show that x^n-bx^2+c has a multiple root if (n^n)(c^n-2)=(4b^n)((n-2)^(n-2))
b)Show co4@=8(cos@-cosPI/8)(cos@-cos3PI/8)(cos@-cos5PI/8)(cos@-cos7PI/8)
f(x)=x^n-bx^2+c
f'(x)=nxn-1-2bx
=x(nxn-2-2b)
=0 when nxn-2-2b=0 x=/=0
xn-2=2b/n
x=(2b/n)1/(n-2) (that is the condition for multiple root)

sub it back into f(x).
f((2b/n)1/(n-2))=(2b/n)n/(n-2)-b(2b/n)2/(n-2)+c
=(2b/n)2/(n-2)((2b/n)-b)+c=0
(2b/n)(2b/n-b)n-2=-cn-2

probably did something wrong

b)Show co4@=8(cos@-cosPI/8)(cos@-cos3PI/8)(cos@-cos5PI/8)(cos@-cos7PI/8)
RHS=8(cos@-cos pi/8)(cos@+cos pi/8)(cos@-3pi/8)(cos@+3pi/8)
=8(cos^2 @-cos^2 pi/8)(cos^2 @-cos^2 3pi/8)
=8(cos^4 @-cos^2 @cos^2 3pi/8-cos^2 pi/8 cos^2 @ + cos^2 pi/8 cos^2 3pi/8)
=8cos^4 @-8cos^2 @(cos^2 3pi/8+cos^2 pi/8)+8cos^2 pi/8 cos^2 3pi/8
=8cos^4 @- 8cos^2 @+8(0.5(cos pi/2+cos pi/4))2
=8cos^4 @ +8cos^2 @ + 1

LHS=cos 4@=2cos^2 2@-1
=2(2cos^2 @-1)^2 -1
=2(4cos^4 @ -4cos^2 @+1)-1
=8cos^4@-8cos^2 @+1=RHS
 
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azureus88

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Lol, i was doing the same question this morning.

For 1a), continuing on from tim, we have [maths]x=(\frac{2b}{n})^\frac{1}{n-2} [/maths] as the root.

[maths]x^n-bx^2+c=0\\x^2(x^{n-2}-b)+c=0\\(\frac{2b}{n})^{\frac{2}{n-2}}(\frac{2b}{n}-b)=-c\\(\frac{2b}{n})^{\frac{2}{n-2}}=\frac{cn}{b(n-2)}\\(\frac{2b}{n})^2=(\frac{cn}{b(n-2)})^{n-2}\\n^2n^{n-2}c^{n-2}=4b^2b^{n-2}(n-2)^{n-2}\\\therefore n^nc^{n-2}=4b^n(n-2)^{n-2}[/maths]
 

Timothy.Siu

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Lol, i was doing the same question this morning.

For 1a), continuing on from tim, we have [maths]x=(\frac{2b}{n})^\frac{1}{n-2} [/maths] as the root.

[maths]x^n-bx^2+c=0\\x^2(x^{n-2}-b)+c=0\\(\frac{2b}{n})^{\frac{2}{n-2}}(\frac{2b}{n}-b)=-c\\(\frac{2b}{n})^{\frac{2}{n-2}}=\frac{cn}{b(n-2)}\\(\frac{2b}{n})^2=(\frac{cn}{b(n-2)})^{n-2}\\n^2n^{n-2}c^{n-2}=4b^2b^{n-2}(n-2)^{n-2}\\\therefore n^nc^{n-2}=4b^n(n-2)^{n-2}[/maths]
good work! too hard for me lol
 

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