Let a,b, and c be real #s with a not equal zero. Find error in the following proof that (-b/2a) is solution to ax^2+bx+c=0.
Let x and y be solutions to the equation. Subtracting ay^2 + by + c = 0 from ax^2 + bx + c = 0 yields a(x^2 + y^2) + b(x -y)=0, which we rewrite as a(x+y)(x-y)+b(x-y)=0. Hence a(x+y)+b=0, and thus x+y=-b/a. Since x and y can be any solutions, we can apply this computation letting y have the same value as x. With y=x, we obtain 2x=-b/a, or x=-b/2a.
2007-01-30
13:08:47
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4 answers
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ClooneyIsAGenius
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Mathematics