Question 8

If $$\alpha, \beta$$ are the roots of the equation $$x^2 + 3x - 3$$ , then the value of $$(\alpha + 1)^{-1} + (\beta +1)^{-1}$$ is equal to

Solution

$$\frac{1}{\alpha\ +1}+\frac{1}{\beta\ +1}=\frac{\left(\alpha\ +\beta\ +2\right)}{\alpha\ \beta\ +\alpha\ +\beta\ +1}$$

Since $$\alpha\ ,\beta\ $$ are roots of the given equation:

$$\alpha+\beta\ =-3$$

$$\alpha\times\beta\ =-3$$

Then we have: $$\frac{\left(\alpha\ +\beta\ +2\right)}{\alpha\ \beta\ +\alpha\ +\beta\ +1}=\frac{\left(-3+2\right)}{-3-3+1}=\frac{1}{5}$$

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