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confused?
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#1
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My exams finished later than anyone else in my school so i missed a few maths lessons. i need help on this question which i know nothing about, just to get me going.


By using substitution, find the equation whose roots are given by adding 2 to the roots of the equation

3x^3 - 13x^2 - 51x - 36 = 0

thanx
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Jonny W
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#2
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Shifting the graph

y = 3x^3 - 13x^2 - 51x - 36

rightwards by 2 units gives

y = 3(x - 2)^3 - 13(x - 2)^2 - 51(x - 2) - 36

So the required equation is

3(x - 2)^3 - 13(x - 2)^2 - 51(x - 2) - 36 = 0
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confused?
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what about this one? this is more simple.

the roots of x^2 - 3x + 1 = 0 are a and b. find the equation whose roots are a+1 and b+1, without finding the values of a and b.
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evariste
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(Original post by confused?)
what about this one? this is more simple.

the roots of x^2 - 3x + 1 = 0 are a and b. find the equation whose roots are a+1 and b+1, without finding the values of a and b.
a+b=3 ab=1
if roots of new eqn are a+1,b+1
sum of roots = a+b+2=5
product of roots=(a+1)(b+1)=ab+1+a+b=1+3+1= 5
so eqn is
x^2-5x+5=0
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BCHL85
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(Original post by confused?)
what about this one? this is more simple.

the roots of x^2 - 3x + 1 = 0 are a and b. find the equation whose roots are a+1 and b+1, without finding the values of a and b.
It likes that
f(x) = 0 has 2 roots a, b
So f(x-1) = 0 will have 2 roots a+1 and b +1
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confused?
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thanx, im getting the hang of it.

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

how would you do this cubic
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BCHL85
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(Original post by confused?)
thanx, im getting the hang of it.

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

how would you do this cubic
Still like that.
f(x) = 0 has 3 roots a, b , c
So f(x-1) = 0 has 3 roots a+1, b+1, c+1
and f(x-1) = 32(x-1)^3 - 14(x-1) + 3
-> that gives you the equation.
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confused?
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(Original post by BCHL85)
Still like that.
f(x) = 0 has 3 roots a, b , c
So f(x-1) = 0 has 3 roots a+1, b+1, c+1
and f(x-1) = 32(x-1)^3 - 14(x-1) + 3
-> that gives you the equation.
am i supposed to substitute.
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BCHL85
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(Original post by confused?)
am i supposed to substitute.
Hmm, it's a long way. What if they give you a question like x^5 + 4x^4 +7x^3 +...???
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confused?
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(Original post by BCHL85)
Hmm, it's a long way. What if they give you a question like x^5 + 4x^4 +7x^3 +...???
im only doing p4. it only goes up to cubic equations. do you know the method p4 students should be using
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BCHL85
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(Original post by confused?)
im only doing p4. it only goes up to cubic equations. do you know the method p4 students should be using
Sorry, Im doing P4 as well. I think you should know it from P2 or P3. As I remember it likes sketching a graph of f(x+a) if you knew graph of f(x)
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confused?
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(Original post by BCHL85)
Still like that.
f(x) = 0 has 3 roots a, b , c
So f(x-1) = 0 has 3 roots a+1, b+1, c+1
and f(x-1) = 32(x-1)^3 - 14(x-1) + 3
-> that gives you the equation.

i still dont get it mate, sorry. How wold you solve this??

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

:confused:
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confused?
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#13
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(Original post by confused?)
i still dont get it mate, sorry. How wold you solve this??

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

:confused:

why are the roots a+1 when it says twice another root?
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BCHL85
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(Original post by confused?)
why are the roots a+1 when it says twice another root?
Sorry to make you more confused . I actually didn't read the question
So if f(x) = 0 has 3 roots a, b, c, f(x/2) = 0 will have 3 roots 2a, 2b, 2c.
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RichE
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#15
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(Original post by confused?)
thanx, im getting the hang of it.

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

how would you do this cubic
If you call the roots a, 2a and b then the cubic factorises as

32(x-a)(x-2a)(x-b) = 32(x^3 - (3a+b)x^2 + (2a^2+3ab)x - 2a^2b).

Comparing the x^2 coefficients gives

3a+b = 0.

Comparing constant coefficients gives

-64a^2 b = 3

Putting b = - 3a gives

a^3 = 1/64

So a= 1/4 and b = -3/4.

Just to be on the safe side we can check

2a^2+3ab = 2/16 - 9/16 = -7/16 = -14/32 as required.

So the roots are 1/4, 1/2, -3/4 and the original cubic factors as

(4x-1)(2x-1)(4x+3)
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