Question

Please Solve This....​

Answer 1

Solution:-

$\rm \: {x}^{2} - 5x + 4 = 0$

$\rm \: sum \: of \: the \: zeroes \: = - \dfrac{cofficient \: of \: x}{cofficient \: of \: x {}^{2} }$

$\rm \: product \: of \: zeroes \: = \dfrac{constant \: term \: }{cofficient \: of \: x {}^{2} }$

$\alpha + \beta = \dfrac{ - ( - 5)}{1}$

$\alpha \beta = \dfrac{4}{1}$

$\rm \: \alpha + \beta = 5$

$\alpha \beta = 4$

Now take

$\rm \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } - 2 \alpha \beta$

Now simply

$\rm \: \dfrac{ \beta + \alpha }{ \alpha \beta } - 2 \alpha \beta$

Put the value

$\dfrac{5}{4} - 2 \times 4$

$\rm \: \dfrac{5}{4} - 8$

$\dfrac{5 - 8 \times 4}{4}$

$\dfrac{5 - 32}{4}$

$\to \dfrac{27}{4}$

Answer 2

Step-by-step explanation:

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