Question

If alpha and beta are the zero of quadratic polynomial ax²+bx+c find 1 alpha +1/beta

Answer 1

Alpha and beta are two the roots of quadratic ax²+bx+c=0

So, alpha + beta = -b/a —(1)

alpha*beta = c/a. — (2)

To find 1/aplha + 1/beta

We get , alpha + beta/alpha*beta , using 1 and 2

= -b/a / c/a

= -b/c = answer

Answer 2

$\huge {\red{\boxed{ \overline{ \underline{ \mid\mathfrak{An}{\mathrm{sw}{ \sf{er}} \colon\mid}}}}}}$

_______________________________

➠ Given :

Roots of equation ax² + bx + c are alpha(α) and beta(β) respectively.

_______________________________

➠ To Find :

We have to find the value of $\sf{\frac{1}{\alpha} + \frac{1}{\beta}}$

_______________________________

➠ Solution :

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

$\sf{\implies \frac{\alpha + \beta}{\alpha \beta}}$

$\bf{We \: know \: that,} \\ \\ \sf{\implies \alpha + \beta = \frac{-b}{a}.......(1)} \\ \\ \sf{\implies \alpha \times \beta = \frac{c}{a}........(2)} \\ \\ \bf{From \: equation \: 1 \: and \: 2} \\ \\ \sf{\implies \frac{1}{\alpha} + \frac{1}{\beta} = \dfrac{\dfrac{\dfrac{-b}{\cancel{a}}}{c}}{\cancel{a}}} \\ \\ \sf{\implies \frac{1}{\alpha} + \frac{1}{\beta} = \frac{-b}{c}}$

_________________________

➠ Additional information :

For quadratic polynomial

$\large{\sf{\alpha + \beta = \frac{-b}{a}}} \\ \\ \sf{\alpha \beta = \frac{c}{a}}$

For cubic polynomial

$\large{\sf{\alpha + \beta + y = \frac{-b}{a}}} \\ \\ \sf{\alpha \beta + \beta y + \alpha y = \frac{c}{a}} \\ \\ \sf{\alpha \beta y = \frac{-d}{a}}$