Question

if alpha and beeta are the zeroes of quadratic polynomial f(x) x2 + x -2, then find the value of 1/alpha-1/beeta​

Answer 1

Answer:

p(x)=x^2+x-2

1/a+1/b=a-b/ab.

=-(a+b)/ab (1)

a+b=-1. (2)

ab=-2. (3)

substituting (2) and (3) in (1) we get

-(-1)/2=1/2

hope this was helpful

Mark me brainlist

Answer 2

Answer:

$given \\ \alpha \: and \: \beta \: are \: the \: zeroes \: of \: f(x) \\ f(x) = {x}^{2} + x - 2 = 0\\ = {x }^{2} + 2x - x - 2 \\ = x(x + 2) - 1(x + 2) \\ = (x + 2)(x - 1) \\ \\ \\ (x + 2) = 0 \\ = x = - 2 \\ \\ \\ (x - 1) = 0 \\ = x = 1 \\ so \: \alpha = ( - 2) \: \: \beta = 1 \\ \\ \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{1}{( - 2)} - \frac{1}{1} \\ = \frac{( - 1) - 2}{2 } \\ = \frac{ - 3}{2}$

Follow this.