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Math Help - modulus

  1. #1
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    modulus

    solve |x^2-16|=|2x-1|
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  2. #2
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    Square both sides:

    (x^2 - 16)^2 = (2x-1)^2

    x^4 - 32x^2 + 256 = 4x^2 - 4x + 1

    x^4 - 36x^2 + 4x + 255 = 0

    You now need to solve this. If you can picture the graphs, you'll know that there is one solution (the most positive) when equating x^2 - 16 = 2x - 1

    x^2 - 2x - 15 = 0

    (x-5)(x+3) = 0

    So, (x-5) and (x+3) are possible roots of the polynomial above. Factoring confirms this.

    x^4 - 36x^2 + 4x + 255 = 0

    (x-5)(x^3 +5x^2 -11x-51)= 0

    (x-5)(x^3 +5x^2 -11x-51)= 0

    (x-5)(x+3)(x^2+2x-17) = 0

    You have a quadratic and two solutions. You can finish this off using the quadratic formula to get the other two remaining roots
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  3. #3
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    Hello, Punch!

    \text{Solve: }\;|x^2-16| \:=\:|2x-1|

    Square both sides: . (x^2-16)^2 \:=\:(2x-1)^2

    . . \text{We have: }\;\underbrace{(x^2-16)^2 - (2x-1)^2}_{\text{difference of squares}} \:=\:0


    Factor: . \bigg[(x^2-16) - (2x -1)\bigg]\,\bigg[(x^2-16) + (2x-1)\bigg] \:=\:0

    . . \text{We have: }\;(x^2-2x-15)(x^2+2x-17) \:=\:0


    x^2-2x-15 \:=\:0\quad\Rightarrow\quad (x-5)(x+3) \:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:5,\:-3}

    x^2+2x-17 \:=\:0 \quad\Rightarrow\quad x \:=\:\dfrac{\text{-}2 \pm\sqrt{72}}{2} \quad\Rightarrow\quad \boxed{x \:=\:\text{-}1 \pm 3\sqrt{2}}


    We squared the equation.
    . . So we must check for extraneous roots.

    But this time, all three roots are acceptable!
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