Question

if ALFA,BETA,GAMA are the zeroes of x^3+3x^2-x-3. Then find the value of 1/alfa+ 1/beta + 1/Gama​

Answer 1

Answer:

-1÷3

Step-by-step explanation:

sum of the roots = -3

product of sum of the roots = -1

product of the roots = 1

Answer 2

Answer:

${x}^{3} + 3 {x}^{2} - x - 3 = 0 \\ > {x}^{2} (x + 3) - 1(x + 3) = 0 \\ > (x + 3)( {x ^{2} - 1)} = 0 \\ > (x + 3)(x + 1)(x - 1) = 0$

so,

$x(1) = \alpha = - 3 \\ x(2) = \beta = - 1 \\ x(3) = \gamma = 1$

so,

$\frac{1 }{ \alpha } = \frac{1}{ - 3} = - \frac{1}{3} = - 0.3333 \\ \frac{1}{ \beta } = \frac{1}{ - 1} = - \frac{1}{1} = - 1 \\ \frac{1}{ \gamma } = \frac{1}{1} = 1$

so sum of the numbers

= -0.3333-1+1

= -0.3333