Question

Let $\alpha, \beta$ be two roots of the equation $x^2 + (20)^{1/4}x + (5)^{1/2} = 0$. Then $\alpha^8 + \beta^8$ is equal to :

• A. 100
• B. 10
• C. 50
• D. 160

Hint:

For high powers of roots, always reduce stepwise: \[ \alpha^2+\beta^2 \rightarrow \alpha^4+\beta^4 \rightarrow \alpha^8+\beta^8 \] If $\alpha^2+\beta^2 = 0$, calculations simplify drastically.

Answer 1

The Correct Option is C

The given quadratic equation is \[ x^2 + 20^{1/4}x + \sqrt{5} = 0 \] Let its roots be $\alpha$ and $\beta$. 
Step 1: Use Vieta’s formulas \[ \alpha + \beta = -20^{1/4}, \qquad \alpha\beta = \sqrt{5} \] Step 2: Find $\alpha^2 + \beta^2$ \[ \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \] \[ = (20^{1/4})^2 - 2\sqrt{5} = \sqrt{20} - 2\sqrt{5} \] Since $\sqrt{20} = 2\sqrt{5}$, \[ \alpha^2 + \beta^2 = 0 \] Step 3: Find $\alpha^4 + \beta^4$ \[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 \] \[ = 0 - 2(\sqrt{5})^2 = -10 \] Step 4: Find $\alpha^8 + \beta^8$ \[ \alpha^8 + \beta^8 = (\alpha^4 + \beta^4)^2 - 2(\alpha\beta)^4 \] \[ = (-10)^2 - 2(5^2) = 100 - 50 = 50 \] \[ \boxed{\alpha^8 + \beta^8 = 50} \]