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flycherrypie

  • 4 years ago

solve (x-2)(x+1)=5 step by step please

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  1. Tomas.A
    • 4 years ago
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    \[(x+2)(x+1)=5\] \[x^2+x+2x+2-5=0\] \[x^2+3x-3=0\] \[D=3\cdot3-4\cdot(-3)\cdot1=21\] \[x_{1,2}=\frac{-3\pm\sqrt{21}}{2}\]

  2. alfie
    • 4 years ago
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    Anyhow, this is how you do a product... (a+b)(c+d) = a*c+b*d+a*d+b*c.

  3. davidmckenna
    • 4 years ago
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    x * x = x^2 x * 1 = 1x -2 * x = -2x -2 * 1 = -2 Then add them all up and simplify where necessary: x^2-x-7 = 0 Then: x^2-x = 7 And from here I'm stuck aslo :L

  4. myininaya
    • 4 years ago
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    thomas i think there was suppose to be a minus between x and 2 \[(x-2)(x+1)=5\] \[x^2-2x+x-2=5\] \[x^2-x-2-5=0\] \[x^2-x-7=0\] \[x=\frac{1 \pm \sqrt{1-4(1)(-7)}}{2}\]

  5. Tomas.A
    • 4 years ago
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    @myininaya oh yeah sorry, i missed it ><

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