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If p(x) = ax2 + bx +c and a + c = b, then one of the zeroes is

b/a

c/a

-c/a

-b/a

$Given a x^{2} + b x + c a n d a + c = b Let α and β be the zeroes of the polynomial \Rightarrow α + β = \frac{- b}{a} - - - - (i) α β = \frac{c}{a} \Rightarrow (α - β)^{2} = (α + β)^{2} - 4 α β = (\frac{- b}{a})^{2} - 4 (\frac{c}{a}) = \frac{b^{2} - 4 a c}{a^{2}} = \frac{(a + c)^{2} - 4 a c}{a^{2}} = \frac{(a - c)^{2}}{a^{2}} \Rightarrow α - β = \frac{a - c}{a} - - - (i i) By solving (i) and (ii) we get α = \frac{c}{a} One of the zero of the given polynomial is \frac{c}{a}$

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