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Math Help - Solve f (x) = 0

  1. #1
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    Solve f (x) = 0

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

    solve f(x) = 0


    if we factorise, we get

    f (x) = 2 (x-3) (x+2) (x^2 5x + 11)

    0 = 2 (x-3) (x+2) (x^2 5x + 11)

    is it meaning that x = 3, and x = 2 ?? or ?
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  2. #2
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    Re: Solve f (x) = 0

    Quote Originally Posted by gilagila View Post
    f (x) = 2 (x-3) (x+2)^2 + 2 (x+2) (x-3)^3

    solve f(x) = 0


    if we factorise, we get

    f (x) = 2 (x-3) (x+2) (x^2 5x + 11)

    0 = 2 (x-3) (x+2) (x^2 5x + 11)

    is it meaning that x = 3, and x = 2 ?? or ?
    Have you written this equation correctly? Is it \displaystyle f(x) = 2(x - 3)(x + 2)2 + 2(x + 2)(x - 3)3 or is it something else? Because if you factorise this, you should get \displaystyle f(x) = 2(x - 3)(x + 2)(2 + 3) = 2(x - 3)(x + 2)5 = 10(x - 3)(x + 2)...
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  3. #3
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    Re: Solve f (x) = 0

    Quote Originally Posted by Prove It View Post
    Have you written this equation correctly? Is it \displaystyle f(x) = 2(x - 3)(x + 2)2 + 2(x + 2)(x - 3)3 or is it something else? Because if you factorise this, you should get \displaystyle f(x) = 2(x - 3)(x + 2)(2 + 3) = 2(x - 3)(x + 2)5 = 10(x - 3)(x + 2)...
    sorry , due to not sure how to use the quadratic and cubic sign, just change it to ^2 and ^3.
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  4. #4
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    Re: Solve f (x) = 0

    One of the things I am sure yuo have already learned is that if abc= 0 then at least one of a, b, or c must be 0.
    Yes, you can factor 2(x-3)(x+2)^2+ 2(x+2)(x-3)^3= 2(x+2)(x- 3)[(x+2)+ (x- 3)^2]= 2(x+2)(x-3)(x^2- 5x+ 11)= 0.

    So you know immediately that x+2= 0, or x- 3= 0, or x^2- 5x+ 11= 0 is 0. If x- 3= 0, then x= 3 so x= 3 is a root. If x+ 2= 0 then x= -2 (NOT 2) so x= -2 is a root. But there are two other roots- numbers that satisfy x^2- 5x+ 11= 0. That does not have any real number roots but you can find the complex roots by using the quadratic formula.
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  5. #5
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    Re: Solve f (x) = 0

    Quote Originally Posted by HallsofIvy View Post
    One of the things I am sure yuo have already learned is that if abc= 0 then at least one of a, b, or c must be 0.
    Yes, you can factor 2(x-3)(x+2)^2+ 2(x+2)(x-3)^3= 2(x+2)(x- 3)[(x+2)+ (x- 3)^2]= 2(x+2)(x-3)(x^2- 5x+ 11)= 0.

    So you know immediately that x+2= 0, or x- 3= 0, or x^2- 5x+ 11= 0 is 0. If x- 3= 0, then x= 3 so x= 3 is a root. If x+ 2= 0 then x= -2 (NOT 2) so x= -2 is a root. But there are two other roots- numbers that satisfy x^2- 5x+ 11= 0. That does not have any real number roots but you can find the complex roots by using the quadratic formula.
    Is it meaning that the x = 3, -2 and (5+√(-19))/2 or (5- √(-19))/2 <- using formula calculation ?
    meaning if f(x) = 0, then x = { 3, 2, (5+√(-19))/2 and (5- √(-19))/2}
    am i right ?
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  6. #6
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    Re: Solve f (x) = 0

    Quote Originally Posted by gilagila View Post
    Is it meaning that the x = 3, -2 and (5+√(-19))/2 or (5- √(-19))/2 <- using formula calculation ?
    meaning if f(x) = 0, then x = { 3, 2, (5+√(-19))/2 and (5- &√(-19))/2}
    am i right ?
    Both x = (5+√(-19))/2 and x = (5- √(-19))/2 are solutions, although, as HallsofIvy mentioned, both of those are complex numbers. That's because x^2-5x+11>0 for all real number values of x. To see that, graph y=x^2-5x+11\,.
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  7. #7
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    Re: Solve f (x) = 0

    Quote Originally Posted by SammyS View Post
    Both x = (5+√(-19))/2 and x = (5- √(-19))/2 are solutions, although, as HallsofIvy mentioned, both of those are complex numbers. That's because x^2-5x+11>0 for all real number values of x. To see that, graph y=x^2-5x+11\,.
    ok. Thanks for the reply
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