Question

if alpha and beta are the two roots of 2x²-6x +3=0,then the value of(alpha/beta+beta/alpha)+ 3(1/alpha+1/beta)+2 alpha beta is​

Answer 1

$\underline\mathfrak{Given:-}$

• alpha and beta are the root of 2x² - 6x + 3 = 0

$\underline\mathfrak{To \: \: Find:-}$

• Find the value of (alpha/beta + beta/alpha) + 3(1/alpha + 1/beta) + 2.alpha.beta ......?

$\underline\mathfrak{Solutions:-}$

2x² - 6x + 3 = 0

• a = 2
• b = -6
• c = 3

$\: \: \: \: \: \therefore {Sum \: \: of \: \: zeroes} \: \: = \: \: \frac{ \: - \: coefficient \: \: of \: \: x}{coefficient \: \: of \: \: {x}^{2}}$

$\: \: \: \: \: \leadsto \: \: {\alpha} \: + \: {\beta} \: \: = \: \: \frac{-b}{a}$

$\: \: \: \: \: \leadsto \: \: {\alpha} \: + \: {\beta} \: \: = \: \: - \: \frac{(-6)}{2}$

$\: \: \: \: \: \leadsto \: \: {\alpha} \: + \: {\beta} \: \: = \: \: \cancel{\frac{6}{2}}$

$\: \: \: \: \: \leadsto \: \: {\alpha} \: + \: {\beta} \: \: = \: \: {3}$

$\: \: \: \: \: \therefore {Product \: \: of \: \: zeroes} \: \: = \: \: \frac{constant \: \: term}{coefficient \: \: of \: \: {x}^{2}}$

$\: \: \: \: \: \leadsto \: \: {\alpha} \: {\beta} \: \: = \: \: \frac{c}{a}$

$\: \: \: \: \: \leadsto \: \: {\alpha} \: {\beta} \: \: = \: \: {\frac{3}{2}}$

Now,

$\: \: \: \: \: \leadsto \: \: {(\frac{\alpha}{\beta} \: + \frac{\beta}{\alpha})} \: + \: {3}{(\frac{1}{\alpha} \: + \frac{1}{\beta})} \: + \: {2} \: \alpha\beta$

$\: \: \: \: \: \leadsto \: \: \frac{{\alpha}^{2} \: - \: {\beta}^{2}}{\alpha\beta} \: + \: {3} \: \frac{({\alpha} \: + {\beta})}{\alpha\beta} \: + \: {2} \: \alpha\beta$

$\: \: \: \: \: \leadsto \: \: \frac{{3}^{2} \: - \: {1}. \: \frac{1}{2}}{\frac{3}{2}} \: + \: \frac{{3}{(3)}}{\frac{3}{2}} \: + \: {2} \: {(\frac{3}{2})}$

$\: \: \: \: \: \leadsto \: \: \frac{{9} \: - \: {3}}{\frac{3}{2}} \: + \: \frac{9}{\frac{3}{2}} \: + \: \frac{6}{2}$

$\: \: \: \: \: \leadsto \: \: \frac{6}{\frac{3}{2}} \: + \: \frac{9}{\frac{3}{2}} \: + \: {3}$

$\: \: \: \: \: \leadsto \: \: \frac{{6} \: \times \: {2}}{3} \: + \: \frac{{9} \: \times \: {2}}{3} \: + \: {3}$

$\: \: \: \: \: \leadsto \: \: \cancel{\frac{12}{3}} \: + \: \cancel{\frac{18}{3}} \: + \: {3}$

$\: \: \: \: \: \leadsto \: \: {4} \: + \: {6} \: + \: {3}$

$\: \: \: \: \: \leadsto \: \: {13}$

Hence,

$\: \: The \: \: value \: \: of \: \: {(\frac{\alpha}{\beta} \: + \frac{\beta}{\alpha})} \: + \: {3}{(\frac{1}{\alpha} \: + \frac{1}{\beta})} \: + \: {2} \: \alpha\beta \: \: is \: \: {13}.$

Answer 2

$\large\mathcal{\green{\underbrace{\orange{SOLUTION:-}}}}$

GIVEN :-

☞ $\rm{{\red{\alpha}}\:and\:{\red{\beta}}}$ are the two roots of the quadratic polynomial “2x² - 6x + 3 = 0” .

CALCULATION :-

✍️ Here,

• coefficient of ‘x²’ = 2

• coefficient of ‘x’ = -6

• constant term = 3

✍️ We have know that,

$\red\bigstar\:\rm{\red{\boxed{\purple{Sum\:of\:the\:roots\:=\:\dfrac{-\:(coefficient\:of\:x)}{coefficient\:of\:x^2}\:}}}}$

$\rm{\implies\:\alpha\:+\:\beta\:=\:\dfrac{-\:(-6)\:}{2}\:}$

$\rm{\implies\:\alpha\:+\:\beta\:=\:\dfrac{6}{2}\:}$

$\rm{\blue{\implies\:\alpha\:+\:\beta\:=\:3\:}}$

✍️ Also,

$\red\bigstar\:\rm{\red{\boxed{\purple{Product\:of\:the\:roots\:=\:\dfrac{constant\:term}{coefficient\:of\:x^2}\:}}}}$

$\rm{\blue{\implies\:\alpha\:{\beta}\:=\:\dfrac{3}{2}\:}}$

✍️ Now we calculate the value of

$\rm{\left\{\:\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:\right\}\:\:+\:\:3\:\left\{\:\dfrac{1}{\alpha}\:+\:\dfrac{1}{\beta}\:\right\}\:\:+\:\:2\:\alpha\:\beta\:}$

$\rm{=\:\left\{\:\dfrac{\alpha^{2}\:+\:\beta^{2}}{\alpha\:\beta}\:\right\}\:\:+\:\:3\:\left\{\:\dfrac{\alpha\:+\:\beta}{\alpha\:\beta}\:\right\}\:\:+\:\:2\:\alpha\:\beta\:}$

$\rm{=\:\left\{\:\dfrac{(\alpha\:+\beta)^2\:-\:2\alpha\beta}{\alpha\:\beta}\:\right\}\:\:+\:\:3\:\left\{\:\dfrac{\alpha\:+\:\beta}{\alpha\:\beta}\:\right\}\:\:+\:\:2\:\alpha\:\beta\:}$

$\rm{=\:\left\{\:\dfrac{3^2\:-\:2\times{\dfrac{3}{2}}}{\dfrac{3}{2}}\:\right\}\:\:+\:\:3\:\left\{\:\dfrac{3}{\dfrac{3}{2}}\:\right\}\:\:+\:\:2\times{\dfrac{3}{2}}\:}$

$\rm{=\:\dfrac{2\times{6}}{3}\:+\:3\times{2}\:+\:3\:}$

$\rm{=\:4\:+\:6\:+\:3\:}$

$\rm{\pink{=\:13}}$

$\red\therefore\:\rm{\green{\boxed{\orange{\left\{\:\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:\right\}\:\:+\:\:3\:\left\{\:\dfrac{1}{\alpha}\:+\:\dfrac{1}{\beta}\:\right\}\:\:+\:\:2\:\alpha\:\beta\:\:=\:\:13\:\:}}}}$