Oliver Wood	Posted on Monday, 21 April, 2003 - 06:18 pm: Here is an entire question, the last part of which i can't do: In this question, f(x)=(1-x)^5+(2+x)^5 q.i) use binomial expansions to express f(x) as a polynomial in (x) a.i) x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32 q.ii)show that f'(x)= 15(4x^3 + 6x^2 + 12x +5) a.ii)using differentiation q.iii) show that 2x+1 is a factor of (4x^3...5) a.iii) 2x^2 + 2x + 5 q.iv) It follows that the point with x-coordinate -1/2 is a stationary point on the curve y=f(x). determine whether this point is a maximum, a minimum or a point of inflection. a.iv) minimum q.v) Factorise 4x^3 + 6x^2 + 12x + 5, hence show that the curve y=f(x) has no other stationary point. a.v) so far i have written: 4x^3 + 6x^2 + 12x + 5 = (2x^2 + 2x + 5)(2x+1) i think i need to factorise the (2x^2 + 2x + 5) but i'm not sure how to!! if someone spots a mistake in any of the five parts could they please tell me, or if they know how to finish off the question please tell me. Thank you Oliver
David Loeffler	Posted on Monday, 21 April, 2003 - 06:58 pm: We want to show that there's no other point where 4x3 + 6x2 + 12x + 5 = 0, so we want to show that 2x2 + 2x + 5 is never zero. Do you know about the discriminant of a quadratic? (Alternatively: 2x2 + 2x + 5 = x2 + (x+1)2 + 4 ³ 4, since the square of anything is positive; so this is always at least 4 and consequently never zero. But discriminants provide a more general method.) David
Matthew Buckley	Posted on Monday, 21 April, 2003 - 08:14 pm: Or, if you have met differentiation, then you should know about the formula for solving quadratic equations - can you see how that will help you?
Oliver Wood	Posted on Tuesday, 22 April, 2003 - 06:35 pm: i have done differentiation but i'm still not sure what to do
Matthew Buckley	Posted on Tuesday, 22 April, 2003 - 07:36 pm: OK. Consider an equation of the form $a{x}^2+\mathrm{bx}+c$, where $a$, $b$, and $c$ are constants, and $a\ne 0$. Divide through by $a$ to get $x^2+(b/a)x+(c/a)=0$. This next part is the hardest part to explain, so please do post back if you are unsure about this next bit: Complete the square to get $(x+(b/2a{))}^2-(b^2/4a^2)=-c/a$, which gives $(x+(b/2a{))}^2=(b^2-4\mathrm{ac})/(4a^2)$. Now take the square root of both sides, remembering that this can be positive or negative, to get ${\displaystyle x+b/(2a)=\pm \frac{\sqrt{b^2-4\mathrm{ac}}}{2a},}$ and then, finally, you get what you want: ${\displaystyle x=\frac{-b\pm \sqrt{b^2-4\mathrm{ac}}}{2a},}$ so if you know what $a$, $b$ and $c$ are, you can then find the roots of your equation - can you see how to factorise your expression now? By the way, I'm not sure what level of maths you're at, so if you unsure about any of the above then we will try to help you in a different way, which may be easier to understand.
Oliver Wood	Posted on Tuesday, 22 April, 2003 - 09:35 pm: thanks very much Matthew + David, I am in Lower Sixth studying Maths (Pure 1, Mech 1, Stats 1) AS and Further Maths (Pure 4, D + D, Mech 2). I have done a couple of exercises from a revision guide on completing the square as the MEI book doesn't show how to find roots using completing the square whereas their exams do! I tried doing the formula to 2x^2 + 2x + 5 but the top part of the formula involves square rooting a negative (2^2 - 40) so i wanted to complete the square. i will have a go with the completing the square and get back when i'm done thanks
Oliver Wood	Posted on Tuesday, 22 April, 2003 - 09:57 pm: i have tried completing the square and i keep going wrong somewhere i got (x+0.5)^2 + 2.25=0 then i take the 2.25 from both sides square root both sides and the answer on the right hand side is the square root of a negative which is possible in the number plain but the pure 1 examiners don't want that, i'm not sure where i ahve gone wrong in my working, maybe it's from the very start of the question, but i can't see what is wrong with it. if you see it or can go through the last part please tell me thank you
Matthew Buckley	Posted on Tuesday, 22 April, 2003 - 10:09 pm: Oliver, well done! You are so close!!!! You have correctly stated that the square root bit is negative, and thus does not have a real square root. So think about what this tells you about what the graph looks like on paper. When the quadratic does have real roots, these roots are when the graph cuts the x axis, so if there are no real roots, what does this tell ou about the quadratic you are dealing with? Sorry to be so cryptic all the time, but you will get the most out of this question if you get there with gentle pushes in the right direction, and not by simply being told. Do post back if still stuck - that's what we are here for!
Oliver Wood	Posted on Tuesday, 22 April, 2003 - 10:17 pm: the problem is at pure 1 level no one knows about imaginary numbers so the roots we find have to be real. That's why i think i've done some working wrong, and i understand that as being the best way to get me to learn and i appreciate it, thank you oliver
Matthew Buckley	Posted on Tuesday, 22 April, 2003 - 11:03 pm: OK, so you have not done imaginary roots, so in the world of P1, 2x2 +2x+5 does not have any roots, so its graph never crosse the x axis, however, usually, when you have f(x)=0, you find the roots of the equation (ie, the points given by f(d)=f(e)=0, where d and e are the roots of the equation. However, you have differentiated, and then set your resultant equation equal to 0, so you are trying to find points, say we call the d and e again, where f'(d)=f'(e)=0, ie the roots of f'(x) are WHERE THE TURNING POINTS ARE. You have quite rightly pointed out that 2x2 +2x+5 has no real roots, and thus has no turning points, which gives you the result that you want! Hope that made sense - incidently, you can take this further and differentiate again to find out where f''(x) =0, ie where the second derivative is equal to 0, etc etc. Well done for pursuing the question though.
Oliver Wood	Posted on Wednesday, 23 April, 2003 - 11:11 am: thanks very much, it has helped me greatly