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Maths Qs. Pg 41 Text Test 6

  • 13-09-2014 12:45pm
    #1
    iCrazzy
    Registered Users, Registered Users 2 Posts: 204 ✭✭


    Q.5
    (I)
    Solve the equation:
    x to the power of 4 - 9x squared=0

    (Ii) (2x-1)power of 3(2-x)=0

    Much appreciated.


Comments

  • #2
    Magnate
    Closed Accounts Posts: 877 ✭✭✭Magnate


    What have you tried so far?

    I'll help you with the first and you do the second ;)

    x^4 - 9x^2 = 0

    We can take out an x squared to give:

    x^2(x^2 - 9) = 0

    Then inside the brackets it's a difference of two squares:

    x^2(x+3)(x-3) = 0

    Then that gives us the first point:

    x^2 = 0
    therefore x = 0

    To get the second point

    x + 3 = 0
    x = -3

    And

    x - 3 = 0
    x = 3

    So the answer is x = 0, -3, 3


  • #3
    iCrazzy
    Registered Users, Registered Users 2 Posts: 204 ✭✭iCrazzy


    Magnate wrote: »
    What have you tried so far?

    I'll help you with the first and you do the second ;)

    x^4 - 9x^2 = 0

    We can take out an x squared to give:

    x^2(x^2 - 9) = 0

    Then inside the brackets it's a difference of two squares:

    x^2(x+3)(x-3) = 0

    Then that gives us the first point:

    x^2 = 0
    therefore x = 0

    To get the second point

    x + 3 = 0
    x = -3

    And

    x - 3 = 0
    x = 3

    So the answer is x = 0, -3, 3

    Thank you so much.


  • #4
    iCrazzy
    Registered Users, Registered Users 2 Posts: 204 ✭✭iCrazzy


    Magnate wrote: »
    What have you tried so far?

    I'll help you with the first and you do the second ;)

    x^4 - 9x^2 = 0

    We can take out an x squared to give:

    x^2(x^2 - 9) = 0

    Then inside the brackets it's a difference of two squares:

    x^2(x+3)(x-3) = 0

    Then that gives us the first point:

    x^2 = 0
    therefore x = 0

    To get the second point

    x + 3 = 0
    x = -3

    And

    x - 3 = 0
    x = 3

    So the answer is x = 0, -3, 3

    When I do the second one I get 8x^3-12x^2+6x-1 for the answer of (2x-1)^3
    am i on the right track ?


  • #5
    Magnate
    Closed Accounts Posts: 877 ✭✭✭Magnate


    iCrazzy wrote: »
    When I do the second one I get 8x^3-12x^2+6x-1 for the answer of (2x-1)^3
    am i on the right track ?

    Here you go,

    XU0Xfgj.png


  • #6
    iCrazzy
    Registered Users, Registered Users 2 Posts: 204 ✭✭iCrazzy


    Magnate wrote: »
    Here you go,

    XU0Xfgj.png

    Hi, I am just looking over the question, is there no way you can just solve it out as in remove the brackets by multiplying (2x-1)(2x-1)(2x-1) and then your answer your multiply by (2-x) ??


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  • #7
    Raspberry Fileds
    Closed Accounts Posts: 294 ✭✭Raspberry Fileds


    iCrazzy wrote: »
    Hi, I am just looking over the question, is there no way you can just solve it out as in remove the brackets by multiplying (2x-1)(2x-1)(2x-1) and then your answer your multiply by (2-x) ??

    You'd get: -8x^4 + 28x^3 - 30x^2 + 13x - 2 = 0. How would that help?

    The principle is that, if you multiply two or more unknowns together and the answer is zero, one of those unknowns must be equal to zero.

    Just to clarify something in Magnate's response: when s/he took the cube-root of (2x-1)^3 (which equals 2x-1), s/he was also taking the cube-root of 0 (what's on the other side of the equation) which equals 0.


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