The zeros of the quadratic polynomial p(x) = | vedclass

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Question

(B) Given the quadratic polynomial $p(x) = x^{2} - 3x + 2$.Comparing this with the standard form $ax^{2} + bx + c$,we get $a = 1$,$b = -3$,and $c = 2$

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$\frac{3}{2}$

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(B) Given the quadratic polynomial $p(x) = x^{2} - 3x + 2$.Comparing this with the standard form $ax^{2} + bx + c$,we get $a = 1$,$b = -3$,and $c = 2$.For a quadratic polynomial,the sum of zeros $\alpha + \beta = -\frac{b}{a} = -(\frac{-3}{1}) = 3$.The product of zeros $\alpha \beta = \frac{c}{a} = \frac{2}{1} = 2$.We need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.Taking the common denominator,we get $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$.Substituting the values,$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{2}$.

The zeros of the quadratic polynomial $p(x) = x^{2} - 3x + 2$ are $\alpha$ and $\beta$. Then,$\frac{1}{\alpha} + \frac{1}{\beta} = \ldots$

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