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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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