Question

if a qnd b be the roots of the x²+3x-2=0 then a ÷b+ b÷a+1÷a+1÷b is​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a \: and \: b \: are \: the \: roots \: of \: {x}^{2} + 3x - 2 = 0$

We know,

$\boxed{\purple{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\purple{\bf :\longmapsto\:a + b = - \dfrac{3}{1} = - 3}$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\red{\rm :\longmapsto\:ab = \dfrac{ - 2}{1} = - 2}$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{a}{b} + \dfrac{b}{a} + \dfrac{1}{a} + \dfrac{1}{b}$

$\rm \: = \: \: \dfrac{ {a}^{2} + {b}^{2} }{ab} + \dfrac{b + a}{ab}$

$\rm \: = \: \: \dfrac{ {(a + b)}^{2} - 2ab}{ab} + \dfrac{b + a}{ab}$

$\rm \: = \: \: \dfrac{ {( - 3)}^{2} - 2 \times ( - 2)}{ - 2} + \dfrac{ - 3}{ - 2}$

$\rm \: = \: \: \dfrac{9 + 4}{ - 2} + \dfrac{ 3}{ 2}$

$\rm \: = \: \: \dfrac{ - 13}{2} + \dfrac{ 3}{ 2}$

$\rm \: = \: \: \dfrac{ - 13 + 3}{2}$

$\rm \: = \: \: \dfrac{ - 10}{2}$

$\rm \: = \: \: - 5$

Hence,

$\bf :\longmapsto\:\dfrac{a}{b} + \dfrac{b}{a} + \dfrac{1}{a} + \dfrac{1}{b} = - \: 5$

Additional Information :-

$\rm :\longmapsto\: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta$

$\rm :\longmapsto\: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta( \alpha + \beta )$

$\rm :\longmapsto\: {( \alpha - \beta )}^{2} = {( \alpha + \beta) }^{2} - 4 \alpha \beta$