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anonymous

  • one year ago

If (x+7)(x-2) = ax^2 + bx + c for all values of x, where a, b, and c are constants, what is the value of a + b - c

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  1. amoodarya
    • one year ago
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    expand (x+7)(x-2) then compare rhs ,lhs

  2. UsukiDoll
    • one year ago
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    that's what I was thinking too... if we use the FOIL method to expand (x+7)(x-2) we can figure out what a, b, and c is

  3. amoodarya
    • one year ago
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    \[(x+7)(x-2)=x^2+5x-14=ax^2+bx+c\\ax^2=x^2\\bx=5x\\c=-14\]

  4. amoodarya
    • one year ago
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    another method is : put for example x=0,1,2 both side then you have a system of equation to find a,b,c

  5. anonymous
    • one year ago
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    ohhh okay! It makes much more sense now! Thanks guys!! :)

  6. UsukiDoll
    • one year ago
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    it looked like a simple proof problem where we start from the left side and have it equal to the right side, so by expanding (x+7)(x-2) using FOIL we will have \[x^2+5x-14 \]. Then we compare that equation to \[ax^2+bx+c \] which is the standard quadratic equation.

  7. anonymous
    • one year ago
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    Got it! Thank you!

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