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Prove relation between the zeros and the coefficient of the quadratic polynomial ax^{2} + bx + c

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#### Solution

Let a and b be the zeros of the polynomial ax^{2}+bx +c

`α = (-b+sqrt(b^2-4ac))/(2a)`

`β = (-b-sqrt(b^2-4ac))/(2a) `

By adding (1) and (2), we get

`α + β = (-b+sqrt(b^2-4ac))/(2a)+(-b-sqrt(b^2-4ac))/(2a)`

`=(-2b)/(2a) = (-b)/a = `

Hence, sum of the zeros of the polynomial

`ax^2 + bx + c `

By multiplying (1) and (2), we get

`αβ = (-b+sqrt(b^2-4ac))/(2a)xx(-b-sqrt(b^2-4ac))/(2a)`

`= (b^2-b^2+4ac)/(4a^2)`

`= \frac{4ac}{4a^{2}} = \frac{ c }{ a } `

`= `

Hence, product of zeros = `\frac{ c }{ a }`

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