Question

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


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Answer 1

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

Answer 2

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

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