roots of equations Watch

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sarah12345
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#1
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#1
a cubic has the equation
2x^3 -3x^2 -12x -4=o
and has roots a b and c

find the cubic with integer coeficients which has roots
a-bc, b-ca, and c-ab


thanks alot
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Gaz031
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#2
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#2
(Original post by sarah12345)
a cubic has the equation
2x^3 -3x^2 -12x -4=o
and has roots a b and c

find the cubic with integer coeficients which has roots
a-bc, b-ca, and c-ab


thanks alot
Try evaluating different values of x to find one of the roots to the cubic, then use long division to find the other two roots. Hence find a-bc, b-ca and c-ab.
The cubic you want is of the form (x-a+bc)(x-b+ca)(x-c+ab)=0 [Remember that if f(a)=0 then x-a is a factor of f(x)]
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RichE
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(Original post by sarah12345)
a cubic has the equation
2x^3 -3x^2 -12x -4=o
and has roots a b and c

find the cubic with integer coeficients which has roots
a-bc, b-ca, and c-ab
a+b+c = 3/2
ab + bc + ca = -6
abc = 2

(a-bc) +(b-ca) + (c-ab) = (a+b+c)-(ab+bc+ca) = 3/2 +6 = 13/2

(a-bc)(b-ca) + (c-ab)(a-bc) +(b-ca)(c-ab)

= ab +bc+ ca - (b^2c + a^2c + a^2b+b^2a+c^2a+c^2b) + c^2ab + a^2bc+ b^2ac

= (ab+bc+ca) - (bc+ac+ab)(a+b+c) + 3abc +abc(a+b+c)

= -6 - (-6)(3/2) + 3 x 2 + 2(3/2)

= -6 + 9 + 6 + 3 = 12

I'll leave you to work out

(a-bc)(b-ca)(c-ab)

This is one of the most tiresome maths questions I've ever met - what is the point?!
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sarah12345
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#4
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ha ha yep tis pretty long n pointless!

so to solve these type of equations you work out the sum of the roots and the product?

i unterstand how you got 14/2 and 12 but then how do i use this to work out the equation?


thanks alot
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El Stevo
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#5
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the 13/2 is the sum of the roots = -b/a
the 12 is the sum of the roots in pairs, ie, c/a
the bit riche left you to do is the product of th roots, ie, -d/a

you then rewrite -b/a, c/a, -d/a with the lowest common denominator and then the equation with the new roots pops out....
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Logan
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how about this... factorise x^3-6x^2+9x-4
i got in my notes the solution:
(x-1)(x^2+bx+4)
(x^2+bx^2)
x^2(-1+b)=-6
b=-6+1
b=-5
hence
(x-1)(x^2-5x+4)
(x-1)^2(x-4)

can sum1 explain this to me?
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CorpusNinja
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(Original post by nasht)
how about this... factorise x^3-6x^2+9x-4
i got in my notes the solution:
(x-1)(x^2+bx+4)
(x^2+bx^2)
x^2(-1+b)=-6
b=-6+1
b=-5
hence
(x-1)(x^2-5x+4)
(x-1)^2(x-4)

can sum1 explain this to me?
Yep. You actually have three roots of (x-1)(x-1)(x-4) so the graph cuts the x axis at x=4 but JUST touches it at x=1, hence the double root. Whenever you get a double root, the graph only TOUCHES at that point. If its a triple root then its a point of inflexion. Is that right ppl (plz say yes coz I have two papers tomorrow!!! gah!)
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