Math, asked by harshith9119, 3 months ago

alfha and betha are zeros of ax2+bx+c then find 1/alfha +1/betha


Answers

Answered by rathnakarbillava
0

Answer:

Step-by-step explanation:

ax² + bx +c =0

zeroes α,β

sum of roots = α+β = -b/a

products of roots = αβ = c/a

1/α - 1/β = β-α / αβ   ............(1)

β-a = √(b² - 4ac/a² )      putting this value

1/α - 1/β  = √(b²-4ac/a²)  / c/a   =  √b²-4ac / c

Answered by BloomingBud
7

Correct question:

α and β are zeroes of ax² + bx + c then find  the value of (1/α + 1/β)

Given

  • 'α' and 'β' are two zeroes of polynomial ax² + bx + c.

To find;

  • The value of \boxed{\frac{1}{\alpha}+\frac{1}{\beta}}

So,

Simplifying \frac{1}{\alpha}+\frac{1}{\beta}

= \frac{\beta+\alpha}{\alpha \beta}

Taking LCM as (αβ)

Now,

We know that,

\boxed{\purple{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}

OR

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

And

\boxed{\purple{\tt Product\ of\ the\ zeroes=\frac{c}{a}}}

OR

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

So,

\rightarrow \alpha + \beta = \frac{-b}{a}

\rightarrow \alpha \beta = \frac{c}{a}

Now, putting the respective value, we get,

\boxed{= \dfrac{(\frac{-b}{a})}{\frac{c}{a}}}

= \frac{-b}{\not a} \times \frac{\not a}{c}

= \frac{-b}{c}

Hence,

The value of

\boxed{\frac{1}{\alpha}+\frac{1}{\beta}= \frac{-b}{c}}

Similar questions