Question

If alpha and beta are the zeros of the polynomial f(x) = 5x²
-7x + 1, find the value of 1/alpha+1/beta​

Answer 1

Answer:

5x2+2x-3

by factorising we get,

5x2+5x-3x-3

5x (x+1) - 3(x+1)

(5x-3)* (x+1)

by equating with zero,

5x-3=0, x+1=0

therefore, x=3/5 or x=-1

NOW, 1/alpha + 1/beta

1/3/5+1/-1

5/3+1

5-3/3=2/3

I hope this will help you

Answer 2

Given Equation

$\sf \to5 {x}^{2} - 7x + 1$

To find the value of

$\sf \to \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta }$

Now Take

$\sf \to5 {x}^{2} - 7x + 1 = 0$

Now Compare with

$\sf \to \: a {x}^{2} + bx + c = 0$

We get

$\sf \: a = 5,b = - 7 \: and \: c = 1$

We Know That

$\sf \: \to \: sum \: of \: zeroes( \alpha + \beta ) = \dfrac{ - b}{a}$

$\sf \to \: product \: of \: zeroes( \alpha \beta ) = \dfrac{c}{a}$

Now put the value

$\sf \: \to \: sum \: of \: zeroes( \alpha + \beta ) = \dfrac{ - ( - 7)}{5} = \dfrac{7}{5}$

$\sf \to \: product \: of \: zeroes( \alpha \beta ) = \dfrac{1}{5}$

Now Take

$\sf \to \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = \dfrac{ \alpha + \beta }{ \alpha \beta }$

Put the value

$\sf \to \: \frac{ \dfrac{7}{5} }{ \dfrac{1}{5} } = \dfrac{7}{5} \times \dfrac{5}{1} = 7$

Answer

$\sf \to \: 7$