# Prove relation between the zeros and the coefficient of the quadratic polynomial ax^2 + bx + c - Mathematics

Sum

Prove relation between the zeros and the coefficient of the quadratic polynomial ax2 + bx + c

#### Solution

Let a and b be the zeros of the polynomial ax2+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 }

Concept: Relationship Between Zeroes and Coefficients of a Polynomial
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