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Math Help - Need step-by-step solution

  1. #1
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    Need step-by-step solution

    Solving for roots. I need to explain the procedure to my twelve year old bro who's failing his math class, unfortunately I haven't been able to be of much help, so I'm hoping that you guys can. plus the kind of roots the last three equations have. Thanks

    a) 6(80-30) = 48-x

    b) 2x^2-13x+6 = 0; by factoring

    c) 3x^2 - 24x + 48 = 0

    d) 4x^2 + 7 = 0

    e) 4 (7x-4)^2 +1 = 9
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  2. #2
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    Quote Originally Posted by archistrategos214 View Post
    Solving for roots. I need to explain the procedure to my twelve year old bro who's failing his math class, unfortunately I haven't been able to be of much help, so I'm hoping that you guys can. plus the kind of roots the last three equations have. Thanks

    a) 6(80-30) = 48-x
    Is there something missing here, this is a linear (1st order problem while
    the rest are quadratic (2nd order) problems?

    Any way this goes something like:

    1. Evaluate the left hand side:

    6(80-30) = 6*50 = 300 = 48-x

    so we now have:

    300 = 48-x

    If we subtract 48 from both sides of this equation we still have equality
    so:

    300 - 48 = 48 - x -48

    or:

    252 = -x.

    Now multiply through by -1 to get:

    (-1)*252 = -252 = (-1)(-x) = x,

    or:

    x = -252.

    RonL
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  3. #3
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    Quote Originally Posted by archistrategos214 View Post
    b) 2x^2-13x+6 = 0; by factoring
    We wish to factor this into the form:

    (ax+b)(cx+d) = 2x^2-13x+6 = 0

    Now if this is to be solved by factoring it means that we are looking
    for integers for a, b, c and d.

    So d and d are two different factors of 6 as when we multiply out the
    brackets on the far left the term independent of x is bd (but with both having
    either a + or - sign in front of it, the same sign for both). The factors of 6
    are 1, 2, 3, 6.

    Similarly a and c are two different of 2 as when we multiply out the brackets
    the coefficient of x^2 is ac. The factors of 2 are 1, 2.

    Now trying out all the possibilities we find the one that gives us -13 for the
    coefficient of the x term. The result of this is:

    (2x - 1)(x -6) = 2x^2-13x+6 = 0.

    Now if a product is zero one or other of the terms must be zeros, so
    we have either

    (2x - 1) = 0,

    in which case x = 1/2,

    or

    (x-6) = 0,

    when x = 6.

    So the roots of:

    2x^2-13x+6 = 0

    are x = 1/2 and x = 6.

    RonL
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  4. #4
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    Quote Originally Posted by archistrategos214 View Post
    c) 3x^2 - 24x + 48 = 0
    Divide both sides by 3.
    That means each summand will get divided by three:
    x^2-8x+16=0
    Look at the factors of 16:
    (1,16),(2,8),(4,4)
    We see that using 4,4 when subtracting gives -8 thus,
    (x-4)(x-4)=0
    Make each factor equal to zero:
    x-4=0 and x-4=0
    Thus,
    x=4

    This means any possible solution must be 4 but not necessarily the other way around. We check and indeed see that 4 is a solution to this equation.
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  5. #5
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    Quote Originally Posted by archistrategos214 View Post

    d) 4x^2 + 7 = 0
    Subtract 7 thus,
    4x^2=-7
    Divide by 4 thus,
    x^2=-7/4
    Take square root of both sides,
    x=+\- sqrt(-7/4)=+\- i*sqrt(7)/2


    e) 4 (7x-4)^2 +1 = 9
    Subtract 1,
    4(7x-4)^2=8
    Divide both sides by 4,
    (7x-4)^2=2
    Square root,
    (7x-4)=+\- sqrt(2)
    Add 4,
    7x=4+\- sqrt(2)
    Divide by 7,
    x=(4/7)+\- sqrt(2)/7
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