Question

1. if the roots of (x^2 + 5x - 1 = 0) are (alpha) and (\beta), evaluate the following expressions.

(1) (\frac{1}{alpha^2} + \frac{1}{\beta^2})

(2) (\frac{1}{alpha^3} + \frac{1}{\beta^3})

(3) (\frac{1}{alpha^4} + \frac{1}{\beta^4})

Explanation:

Step1: Find sum and product of roots

Using Vieta's formulas for \(x^2 + 5x - 1 = 0\):

$$\alpha + \beta = -5, \quad \alpha\beta = -1$$

Step2: Evaluate the first expression

Rewrite the expression using common denominators:

$$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$$

Step3: Calculate the first value

Substitute the known values into the expression:

$$\frac{(-5)^2 - 2(-1)}{(-1)^2} = \frac{25 + 2}{1} = 27$$

Step4: Evaluate the second expression

Rewrite the expression using common denominators:

$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^3} = \frac{(\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)}{(\alpha\beta)^3}$$

Step5: Calculate the second value

Substitute the known values into the expression:

$$\frac{(-5)(27 - (-1))}{(-1)^3} = \frac{-5(28)}{-1} = -140$$

Step6: Evaluate the third expression

Rewrite the expression using common denominators:

$$\frac{1}{\alpha^4} + \frac{1}{\beta^4} = \frac{\alpha^4 + \beta^4}{(\alpha\beta)^4} = \frac{(\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2}{(\alpha\beta)^4}$$

Step7: Calculate the third value

Substitute the known values into the expression:

$$\frac{(27)^2 - 2(-1)^2}{(-1)^4} = \frac{729 - 2}{1} = 727$$

Answer:

(1) \(27\)
(2) \(-140\)
(3) \(727\)