C1 question

Traceur

Hi there. How would I go about doing this question below...?

Would appreciate any help please!

Reply 1

Traceur

OP

Can anyone help please?

Reply 2

gozatron

16

Original post by Traceur

Can anyone help please?

Is the answer x^3 - 5x^2 + 3x + 9 ? If so i can help you.

Reply 3

marek35

This is pretty straightforward

Well you know that x =-1 and touches x at 3

Which means you can compile the cubic via the inverse.

(x+1) (x-3)(x-3)

Muliply those out and you'll get: x^3- 5x^2 +3x+9=0

As you can see clearly a=-5

got it?

Reply 4

Traceur

OP

Original post by marek35

This is pretty straightforward

Well you know that x =-1 and touches x at 3

Which means you can compile the cubic via the inverse.

(x+1) (x-3)(x-3)

Muliply those out and you'll get: x^3- 5x^2 +3x+9=0

As you can see clearly a=-5

got it?

Ah of course! Thank you!

Reply 5

marek35

Original post by Traceur

Ah of course! Thank you!

no problem

:biggrin:

Reply 6

Brecon

I'm pretty sure you could also have done it with simultaneous equations as well, although it's slightly more strenuous and long winded! :P

Reply 7

Traceur

OP

Thanks for your answer but it surely seems to be much easier to use mareks solution...
I did think to use simultaneous equations but its unnecessarily complicated (for me anyway).

(edited 13 years ago)

Reply 8

marek35

Original post by Traceur

Thanks for your answer but it surely seems to be much easier to use mareks solution...
I did think to use simultaneous equations but its unnecessarily complicated (for me anyway).

indeed solving it simultaneously would be pointless as you know where it crosses the x axis (or touches in this case too).

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