Question

# If $$\alpha$$ and $$\beta$$ are the zeros of the quadratic polynomial $$f(x)=x^{2}-px+q$$, then find the values of (i) $$\alpha^{2}+\beta^{2}$$ (ii) $$\dfrac {1}{\alpha }+\dfrac{1}{\beta}$$

Solution

## Consider the given polynomial $$f\left( x \right)={{x}^{2}}-px+q$$ $$\alpha$$ and $$\beta$$are the zeros of given polynomial   We know that, $$Sum of zeros$$=$$-\dfrac{cofficent\,of\,x}{cofficent\,of\,{{x}^{2}}}$$ $$\alpha +\beta =\dfrac{-\left( -p \right)}{1}$$ $$\alpha +\beta =p$$ $$Product of zeros=\dfrac{constant}{cofficent\,of\,{{x}^{2}}}$$ $$\alpha \times \beta =\dfrac{q}{1}$$ $$\alpha \times \beta =q$$   Then, Part (1):- $${{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta$$ $${{\alpha }^{2}}+{{\beta }^{2}}={{p}^{2}}-2q$$   Part (2):- $$\dfrac{1}{\alpha }+\dfrac{1}{\beta }=\dfrac{\alpha +\beta }{\alpha \beta }$$ $$=\dfrac{p}{q}$$ Hence, this is the answer.Maths

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