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Some Maths Help

  1. Apr 15, 2004 #1
    I’d really appreciate some help with these problems. I know it seems like I’m probably just asking you to do my entire assignment for me – but I assure you, this is (sadly) just a tiny piece of what I’ve got to do. I’ve attempted all of them, and still not coming up with any solutions. I’d appreciate any help you can give:

    Q.2 (a) x^2 + 2x + p is a factor of x^3 + ax^2 + bx + c. Show that c = (a – 2)(b – 2a + 4).

    What I’ve done so far:
    I divided the (^2 + 2x + p) into (x^3 + ax^2 + bx + c), hoping to be able to equate the remainder to zero and then find c. I got c = 2x^2 + 2ax – bx + px + pa. I don’t know whether this is right (probably isn’t) or what I should do next.

    Q.6 (c) Find the co-ordinates of the minimum point of the function f: x  xe^x – 4e^x.

    I differentiated xe^x – 4e^x and equated it to zero, getting an answer of x=3. I then subbed this back into xe^x – 4e^x and got a co-ordinate of (3 , 3e^3 – 4e^3). This doesn’t seem very right to me… any ideas?

    These two are the only ones I’ve got at the minute, but I’m sure I’ll be back with more after I’ve attempted the rest a few more times.

    PS - Great new forum layout, haven't had a chance to visit since the redesign.
  2. jcsd
  3. Apr 15, 2004 #2
    [tex] x^3+ax^2+bx+c = (x+k)(x^2 +2x +p) [/tex]

    u can compare the coefficient of same degree in x

    u will observe
    k= a-2 and p= b-2a +4

    and c= pk

    u have done the Pro 2 correctly the point is (3,-e3)
  4. Apr 16, 2004 #3


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    "I divided the (^2 + 2x + p) into (x^3 + ax^2 + bx + c), hoping to be able to equate the remainder to zero and then find c. I got c = 2x^2 + 2ax – bx + px + pa."

    That's a good start but your answer for c should not have an "x" in it. Also, notice that the answer you want does not have "p" in it. When you divide (x^2+ 2x+ p) into (x^3+ ax^2+ bx+ c), you get a remainder of (b-p-2a-4)x+ c+ap-2p which has to be 0 FOR ALL x! That means the two coefficients must be 0: b- p- 2a- 4= 0 and c+ ap- 2p= 0. You can use those two equations to eliminate the p: from the first, p= b-2a-4 and substitute that into the second equation. NOW solve for c!

    "I differentiated xe^x – 4e^x and equated it to zero, getting an answer of x=3. I then subbed this back into xe^x – 4e^x and got a co-ordinate of (3 , 3e^3 – 4e^3). This doesn’t seem very right to me… any ideas?"

    My only advice would be to do the arithmetic and finish the problem! 3e^4- 4e^3= what??
  5. Apr 16, 2004 #4
    I must be seriously loosing it when I'm asking for help in simple subtraction...

    Moving swiftly on...
    Ok, so here’s two more I’m having difficulty with:

    Q.1 (B) The circle K has equation x^2 + y^2 – 8x – 8y + 28 = 0. a and b are two points on K such that length of ab = 2root3.

    (I) Write down the co-ordinates of the centre and the radius length of K and hence, find the perpendicular distance from the centre of K to the chord ab.
    (II) Given the the coordinated of a are (2,4) and the slope of ab>0, find the equation of ab.

    Ok, so part (I) is fine, but I’m having trouble with part (II). The centre of the circle is (4,4) and the radius is 2 units. The distance from a to b is 2root3, so using the distance formula and the point a(2,4) and b(x,y), I tried to find the equation. I don’t think this is the right way to go about it.

    (C) C is the circle with equation x^2 + y^2 – 10x – 8y +25 = 0.
    (I) Show that the x-axis is a tangent to C and hence, or otherwise, find the coordinates of p, the point of contact.

    Ok, I'm totally lost here. I found the centre of the circle to be (5,4) and radius 4 units. The perpendicular distance from (5,4) to the x-axis should be 4 unites, but I'm not sure how to prove this. Any advice?

    Thanks in advance.
  6. Apr 16, 2004 #5
    1.b.ii) It is the right way to go about it. But if you apply only the distance formula, you're going to get the equation for a circle about (2,4) with radius 2sqrt(3). Get an equation from the distance formula, and then apply the equation for your original circle to narrow it down to two points. Select the point that yields a line of positive slope.

    1.c.i) What's the derivative of the circle at x = 5?

  7. Apr 17, 2004 #6
    But how exactly do I "apply" the equation to the circle?

    I don't really understand this method at all?
  8. Apr 17, 2004 #7
    It gives you a system of two equations in two variables. You have to use your algebra 2 skills to eliminate one of the variables and solve for the second.

    You're familiar with derivatives, right? You must be, you used them to solve one of the first problems.

    Let me rephrase the question it's asking you. "Find the equation of the tangent line to the circle at the point (5,0)." You'll need to use more algebra 2 skills plus a little derivative (which you really shouldn't need calculus to calculate). You should find that this equation is very easily written...

  9. Apr 18, 2004 #8
    I assume you're reffering to some kind of American curriculum when you refer to "Algebra 2"?
  10. Apr 18, 2004 #9
    I'm assuming "solve a system of equations," right along with "find the equation of a tangent line" means the same thing in all the English speaking world...

  11. Apr 18, 2004 #10
    Well yeah, but when you say "use your Algebra 2 skills" I'm just curious as to what I'm actually supposed to use...
  12. Apr 18, 2004 #11
    Algebra 2 is a second year algebra class in the US. It generally covers more complicated algebraic manipulation, including systems of equations and the point-slope form of linear equations, both of which will be convenient. Algebra 1 basically covers simple algebraic manipulation (solving for a variable using only the arithmetic operators) and the y = mx + b form of linear equations, and briefly polynomial equations of slightly higher order.

  13. Apr 18, 2004 #12
    Ok, so do you mean to solve the two equations simultaneously then? If that's what you mean, I just get an answer of "8x = 36"... Yes, I would also assume that "sole a system of equations" would mean the same thing throughout the english speaking world, however I am not a native english speaker, and have never heard anybody use the phrase "solve a system of equations".
  14. Apr 18, 2004 #13
    There you go.
  15. Apr 18, 2004 #14
    Um, I dont think that refers to my (current) question?
  16. Apr 18, 2004 #15
    Yes, you need to solve the system of simultaneous equations, which means the same thing as solving the system of equations simultaneously.

    But I'm not getting x = 9/2 as the solution. You will get one x value and two y values, though, for two total solutions.

  17. Apr 19, 2004 #16
    Ok, so I've tried it again... getting x=9, with a complex root when I sub that in to get my two y values, which can't be right.
  18. Apr 19, 2004 #17
    You're right, imaginary points of intersection would be bad.

    What are the two equations that you're using? The math actually worked out pretty nicely if I recall, so it seems unlikely you'd make a mistake there twice.

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