Question

If alpha and beta are zeroes of x2 -x-1 find the value of 1/alpha +1/beata

Answer 1

Answer:

$\large\bold\red{-1}$

Step-by-step explanation:

Given,

A quadratic equation,

${x}^{2} - x - 1 = 0$

Also,

$\alpha \: and \: \beta \: are \: the \: zeroes$

Therefore,

We have,

$\alpha + \beta = 1$

And,

$\alpha \beta = - 1$

Now,

We have to find,

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

Further Simplifying,

We get,

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

Putting the respective values,

We get,

$= > \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{1}{ - 1} = - 1 \\ \\ = > \large \boxed{ \bold{ \frac{1}{ \alpha } + \frac{1}{ \beta } = - 1}}$

Answer 2

Question :---- we have to find 1/a + 1/b ?

Given :----

• Equation = x² - x -1
• roots are = a and b ...

Formula to be used :---

• Brahmacharya Formula
• (a²-b²) = (a+b)(a-b)

roots of Equation ax²+bx+c = 0

$\frac{ - b ±\sqrt{ {b}^{2} - 4ac } }{2a}$

Here, a = 1, b = -1 , c = -1 ,

Putting values in formula now we get,

$\frac{ - ( - 1)±\sqrt{1 - 4( - 1)} }{2} \\ \\ \frac{1± \sqrt{5} }{2} \\ \\ so \: its \: roots \: are \: \\ \\ \frac{ \sqrt{5} + 1}{2} \: \: and \: \: \frac{ \sqrt{5} - 1}{2}$

Now, we have to find 1/a + 1/b :----

Putting values again we get,

$\frac{1}{ \frac{ \sqrt{5} + 1}{2}} \: + \frac{1}{ \frac{ \sqrt{5} - 1}{2}} \\ \\ \frac{2}{ \sqrt{5} + 1 } + \frac{2}{\sqrt{5} - 1 } \: \\ \\ \frac{2 \sqrt{5} - 2 + 2 \sqrt{5} + 2 }{( \sqrt{5} + 1)( \sqrt{5} - 1) } \\ \\ \frac{4 \sqrt{5} }{( \sqrt{5})^{2} - 1^{2} } \\ \\ \frac{4 \sqrt{5} }{4} \\ \\ \implies \: \sqrt{5}$

So, our answer will be = √5 .....

(Hope it helps you)