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AS Further Mathematics FP1 Rational Functions question

Hi all,

I need help with 5 (b) (i), please. I have tried to find discriminant (D = b^2 - 4ac) but to no avail.

Please do explain this.

Thank you very much!

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Reply 1

Zacken

Original post by londoncricket

Hi all,

I need help with 5 (b) (i), please. I have tried to find discriminant (D = b^2 - 4ac) but to no avail.

Please do explain this.

Thank you very much!

As you've said we have

\displaystyle (x+1)(x-3) = kx(x-2)

Expanding gives us:

\displaystyle x^2 -2x -3 = kx^2 - 2kx \Rightarrow x^2(k-1) + x(2 - 2k) + 3 = 0

So what's the discriminant of this quadratic?

Reply 2

londoncricket

Thank you for your prompt reply!

The discriminant of this quadratic would be:

4k^2 - 20k + 16

(Sorry, I am not well versed in LaTeX).

Reply 3

Zacken

Original post by londoncricket

Thank you for your prompt reply!

The discriminant of this quadratic would be:

4k^2 - 20k + 16

Now you know that that the discriminant should be

\geq 0

, since the "intersect" means that there is either one or two solutions.

This gives us

4k^2 - 20k + 16 > 0 \Rightarrow 4(k^2 - 5k + 4)> 0 \Rightarrow k^2 - 5k + 4 > 0

. By dividing by

4

.

Could you, perchance - factorise this for me?

(Sorry, I am not well versed in LaTeX).

Don't worry one iota, you're coming across perfectly clear. You do want to press the "reply" button at the bottom-right corner of my post so that I get a notification when you reply to me. :-)

Reply 4

londoncricket

Original post by Zacken

Now you know that that the discriminant should be

\geq 0

, since the "intersect" means that there is either one or two solutions.

This gives us

4k^2 - 20k + 16 > 0 \Rightarrow 4(k^2 - 5k + 4)> 0 \Rightarrow k^2 - 5k + 4 > 0

. By dividing by

4

.

Could you, perchance - factorise this for me?

Don't worry one iota, you're coming across perfectly clear. You do want to press the "reply" button at the bottom-right corner of my post so that I get a notification when you reply to me. :-)

That's great. Thanks very much!

I was getting those answers before, but I did not think to factorise the values derived for k, hence not getting it in the form the question wanted it in.

Thanks a tonne for your help! Definitely deserve a rep for this.