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If

/frac{1}{a}(a+b-c) /frac{1}{a}(a-b-c) /frac{1}{a}(a-b+c) /frac{1}{c}(c-b+a)

Saran .P.S by SARAN .P.S , 20, India

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1 solution

Tony Sprinkle
Jan 5, 2015

a x 2 + b x + c = ( x − α ) ( x − β ) ax^2 + bx + c = (x - \alpha)(x - \beta)

Let:

α = − b + b 2 − 4 a c 2 a and β = − b − b 2 − 4 a c 2 a \alpha =\frac{-b + \sqrt{b^2 - 4ac}}{2a}\quad\quad\text{and}\quad\quad\beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a}

Thus:

1 + α = 2 a − b + b 2 − 4 a c 2 a and 1 + β = 2 a − b − b 2 − 4 a c 2 a 1+ \alpha =\frac{2a - b + \sqrt{b^2 - 4ac}}{2a}\quad\quad\text{and}\quad\quad 1 + \beta = \frac{2a -b - \sqrt{b^2 - 4ac}}{2a}

Multiplying these two expressions together and simplifying gives the answer:

a − b + c a \boxed{\dfrac{a - b + c}{a}}

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