Question

3. If a.ß are the zeroes of the polynomials f(x) = x² + x + 1 , then 1/a + 1/b​

Answer 1

Answer:

2

Step-by-step explanation:

Quadratic formula: $\frac{-b\ +- \sqrt{b^2 - 4ac} }{2a}$

a = $\frac{-1 + \sqrt{1 - 4}}{2}$

b = $\frac{-1 - \sqrt{1 - 4}}{2}$

1/a = 2/$\sqrt{3} - 1$

1/b = 2/-($\sqrt{3} + 1$)

1/a + 1/b = $\frac{2}{\sqrt{3} - 1} + \frac{2}{-\sqrt{3} - 1}$

= $\frac{2[\sqrt{3} - 1 -\sqrt{3} - 1]}{(-1)(\sqrt{3} + 1)(\sqrt{3} - 1)}$

= $\frac{2[\sqrt{3} - 1 -\sqrt{3} - 1]}{-1(3 - 1)}$

= $\frac{2[\sqrt{3} - 1 -\sqrt{3} - 1]}{-2}$

= $-1(\sqrt{3} - 1 - \sqrt{3} - 1)$

= -1(-2)

= 2