Question

if a and b are the zeroes of x^2 + 5x -2 then the value of
$\frac{1}{a} + \frac{1}{b}$
​

Answer 1

${\huge{\boxed{\underline{\underline{\bf Given:-}}}}}$

$\sf \alpha \;\&\;\beta \;are\;zeroes\;of\;x^2+5x-2$

${\huge{\boxed{\underline{\underline{\bf To\;Find:-}}}}}$

$\sf \dfrac{1}{\alpha }+\dfrac{1}\beta$

${\huge{\boxed{\underline{\underline{\bf Solution:-}}}}}$

$\sf We\;know\;that$

$\sf \alpha +\beta =\dfrac{-b}{a}$

$\sf \alpha \beta =\dfrac{c}{a}$

$\sf On\;comparing\;the\;equation\;with\;ax^2+bx+c$

$\sf a=1\;\&\;b=5\;\&\;c=-2$

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

$\sf\alpha +\beta =-5$

$\sf \alpha \beta =\dfrac{-2}{1}$

$\sf \alpha \beta =-2$

$\sf Now,\;Finding$

$\sf\dfrac{1}{\alpha }+\dfrac{1}{\beta }$

$\sf\dfrac{\alpha +\beta }{\alpha \beta }$

$\sf\dfrac{-5}{-2}$

$\sf\dfrac{5}{2}$

Answer 2

❍ Given :-

• α and β are zeroes of x² + 5x - 2

❍ To Find :-

• $\dfrac{1}{ \alpha } + \dfrac{1}{ \beta }$

❍ Solution :-

We know that,

$→\boxed{\sf \alpha + \beta = \dfrac{ - b}{a}}$

$→\boxed{\sf \alpha \beta = \dfrac{c}{a}}$

On comparing the equation with ax² + bx + c,

1. a = 1
2. b = 5
3. c = -2

1•} $\sf \alpha + \beta = \dfrac{ - 5}{1}$

Hence, $\sf \alpha + \beta = - 5$

2•} Now, $\sf \alpha \beta = \dfrac{ - 2}{1}$

Hence,$\sf \alpha \beta = - 2$

➸ Now we have to find $\dfrac{1}{ \alpha } + \dfrac{1}{ \beta }$,

$→ \dfrac{ \alpha + \beta }{ \alpha \beta }$

$\sf → \dfrac{\cancel{-} 5}{\cancel{-} 2}$

$\sf → \dfrac{5}{2}$

Hence, the answer is $\sf \dfrac{5}{2}$.