Original post by subbhy

Hi, could someone help with this question please.

I multiplied by x on both sides, with the condition that x>0 but that didn’t get me far because x>0 wasn’t part of any of the options.

How could I do this?

I probably wouldn’t multiply by x, since x can be negative and this actually makes the inequality problematic to solve as it may or may not flip it.

So since x is problematic to multiply it by, can you think of a multiple of x that is always positive and therefore always keeps the inequality sign the right way round?

Original post by UtterlyUseless69

I probably wouldn’t multiply by x, since x can be negative and this actually makes the inequality problematic to solve as it may or may not flip it.

So since x is problematic to multiply it by, can you think of a multiple of x that is always positive and therefore always keeps the inequality sign the right way round?

So since x is problematic to multiply it by, can you think of a multiple of x that is always positive and therefore always keeps the inequality sign the right way round?

Aha! so we multiply by x^2

And then use the factor theorem?

Original post by subbhy

Aha! so we multiply by x^2

And then use the factor theorem?

And then use the factor theorem?

Indeed. As such, your inequality becomes

x^4 - 6x^3 + 9x^2 - 4x > 0

Now you need to treat it like

x^4 - 6x^3 + 9x^2 - 4x = 0

To find the critical values

You could use the factor theorem, but before doing so, it might be nice to take out a common factor from all the terms first.

(edited 3 weeks ago)

Original post by subbhy

Aha! so we multiply by x^2

And then use the factor theorem?

And then use the factor theorem?

Must admit, Id simply factorise the numerator / sketch the cubic. Then note that the intervals you want are when the numerator and denominator are both negative or both positive. This is even simpler than expected due to the repeated root in the numerator.

Original post by mqb2766

Must admit, Id simply factorise the numerator / sketch the cubic. Then note that the intervals you want are when the numerator and denominator are both negative or both positive. This is even simpler than expected due to the repeated root in the numerator.

Perhaps, though I would recommend they follow the approach suggested previously as it is good practice and examiners at A level (and equivalent) can be quite harsh about stuff like this.

(edited 3 weeks ago)

Original post by UtterlyUseless69

Perhaps, though I would recommend they follow the approach suggested previously as it is good practice and examiners at A level (and equivalent) can be quite harsh about stuff like this.

The thing is that its extra algebra for little reason. Multiplying by x (after multiplying through by x^2 and dividing by x) is no different from dividing by x if youre only interested in the sign/zero of the answer. In addition, theres no reason to write it as a quartic

x^4 - 6x^3 + 9x^2 - 4x > 0

just to factor out the x on the left which youve multiplied it by in the first place. Also being pedantic, getting in the habit of multiplying through by x^2 and not writing that the resulting equation / inequality is not valid when x=0 can give incorrect reasoning at times (not important for this question though). So x=0 is not a root of the original expression as it is not defined at that point, though its where the expression changes sign.

So

(x-1)^2 (x-4) / x > 0

is equivalent to

(x-4) / x > 0

which is true when

x<0 and x<4 (both negative) so x<0

x>0 and x>4 (both positive) so x>4

There is no reason that wouldnt get full marks at a level.

(edited 3 weeks ago)

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