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Math Help?


mydiamond

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I'm sorry to destroy the leisurely Holiday theme, but if you're good at Algebra please do lend a hand :( I can't find my notes on this one particular material and I have no clue what am I doing!

 

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They're due tomorrow :(

 

EDIT: apparently variable 'i' isn't accepted in the final answer. I've tinkered around with these two questions since yesterday and I'm going crazy

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Liv, sorry. I am good at algebra, but this goes beyond my knowledge base.

Huskies in the Heartland

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-5 and 3i are solutions means that (x + 5) and (x - 3i) are factors of the equation.

 

f(x) = (x + 5)(x - 3i)(ax + B)

f(-1) = 120 = (-1 + 5)(-1 - 3i)(-a + B)

 

Multiple this out and you should be able to work out the values for a and b

 

For the next problem, you will have two unknown factors of the equation so you will have two factors (ax + B) and (cx + d) to work out.

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-5 and 3i are solutions means that (x + 5) and (x - 3i) are factors of the equation.

 

f(x) = (x + 5)(x - 3i)(ax + B)

f(-1) = 120 = (-1 + 5)(-1 - 3i)(-a + B)

 

Multiple this out and you should be able to work out the values for a and b

 

For the next problem, you will have two unknown factors of the equation so you will have two factors (ax +  B) and (cx + d) to work out.

 

Thank you thank you thank you! I simply multiplied the (x+5) (x-3i) and forgot the last bit

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EDIT: Sorry, Olivia. I couldn't find the binder. Maybe I threw them out.


Hey, Olivia....I might still my Calculus notes from last year. If it isn't too late, I can try to find the material on how to do it, and then scan you a copy. 

(I knew those Calculus notes would come in handy someday!)

Edited by SolitaryHowl
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I tried to do it using that method and realised there is an easier method.

 

Consider

 

f(x) = A x^3 + B x^2 + C x + D

 

Work out f(3i). This equals 0, so the real and imaginary parts of the equation are both 0. This gives you 2 equations for A, B, C & D.

 

Now work out f(-5) which also equals 0. This gives you another equation for A, B, C & D.

 

Finally work out f(-1) which equals 120. This gives you a 4th equation for A, B, C & D.

 

These 4 equations are sufficient to work out the values for A, B, C & D.

 

 

You can then factor the equation knowing that (x + 5) and (x - 3i) are factors.

 

I have worked this through so I can give you more info if required.

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I've worked through the second question as well and it will work in the same way but you will have A, B, C, D & E. Also, when you come to factor the equation don't try using (x - 6i) and (x + 4i) as this is complicated. Instead, substitute y = x^2 and solve for y. Once you have values for y, then replace x in the equations and solve for x. This does sound a bit complicated but it will be easier once you have the equations (I'm trying not to give you the answer).

 

There were also some big numbers (100s in the first question and 1000s in the second question).

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Rant: I just WISH they don't include i anywhere in this equation *grumble grumble*. What annoys me most is that this is actually an easy question made difficult thanks to Mr. i

 

@Elyse: Whoa I don't even speak Calculus :o thanks though!

 

@Adam: Great idea :o I never thought you can set it up like that. Darn. I left my pencil case in my room at the fourth floor.

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