Question

If alpha and Nita are the zeroes of polynomial 2x^2+3x+5 then find the value of alpha+bita\alphabita

Answer 1

Step-by-step explanation:

heya !!

2X² - 3X - 5

Here,

A = 2 , B = -3 and C = -5

Sum of zeroes = -B/A

Alpha + Beta = 3/2

And,

Product of zeroes = C/A

Alpha × Beta = -5/2

Therefore,

Alpha + Beta / Alpha × Beta = 3/2 / -5/2

=> 3/2 × 2/ - 5

=> - 3/5.

if it helps then mrk me as brainliest

Answer 2

Given:-

• p(x) = 2x² + 3x + 5

• $\alpha$ and $\beta$ are the zeroes of polynomial.

To find:-

• Value of $\sf \dfrac{ \alpha + \beta }{ \alpha \beta}$

Solution:-

★ GivEn Polynomial:- 2x² + 3x + 5

This is a form of ax² + bx + c

Therefore,

• a = 2
• b = 3
• c = 5

We know that,

$\normalsize{\underline{\underline{\sf{\purple{\dag\;Sum\;of\;zeroes:-}}}}}$

$:\implies\sf ( \alpha + \beta$ ) = $\sf \dfrac{-b}{a}$

$\normalsize{\underline{\underline{\sf{\purple{\dag\;Product\;of\;zeroes:-}}}}}$

$:\implies\sf ( \alpha \beta$ ) = $\sf \dfrac{c}{a}$

$\therefore\sf ( \alpha + \beta) = \dfrac{-3}{2} ; \sf ( \alpha + \beta) = \dfrac{5}{2}$

Therefore, the value of $\sf \dfrac{ \alpha + \beta }{ \alpha \beta}$ is -

$:\implies\sf \dfrac{ \frac{-3}{ \cancel{2}}}{ \frac{5}{ \cancel{2}}}$

$:\implies\sf \dfrac{-3}{5}$

$\normalsize{\underline{\underline{\sf{\red{\dag\;Hence\;Solved!!}}}}}$

$\rule{200}{2}$