If the zeroes of the polynomial $x^{2} - b x + a$ are two consecutive odd integers, then $b^{2} - 4 a$ is

-5

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Solution

The correct option is D
4
For, $x^{2} + b x + a Let zeroes be α a n d β α + β = - (- b) = b α β = \frac{a}{1} = a The difference between two consecutive odd integers is 2 α - β = 2 \Rightarrow \sqrt{(α + β)^{2} - 4 α β} = 2 \Rightarrow \sqrt{(b)^{2} - 4 (a)} = 2 \Rightarrow (\sqrt{b^{2} - 4 a})^{2} = (2)^{2} \Rightarrow b^{2} - 4 a = 4 b^{2} - 4 a = 4$
So, correct option is d

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