$$\alpha$$ and $$\beta$$ are the roots of quadratic equation. If $$\alpha + \beta = 8$$ and $$\alpha - \beta = 2\surd5$$, then which of the following equation will have roots $$\alpha^4$$ and $$\beta^4$$?

According to question :

$$2\alpha\ =8+\sqrt{5}\ \ or\ \ \ \alpha=4+\sqrt{5}\ .$$

And, $$2\beta=8-\sqrt{5}\ \ or\ \ \ \beta=4-\sqrt{5}\ .$$

So, $$\alpha^2=\left(4+\sqrt{5}\right)^2=\left(21+8\sqrt{5}\right).$$

And, $$\beta^2=\left(4-\sqrt{5}\right)^2=\left(21-8\sqrt{5}\right).$$

Again ,

$$\alpha^4=\left(\alpha^2\right)^2=\left(21+8\sqrt{5}\right)^2=\left(761+336\sqrt{5}\right).$$

$$\beta^4=\left(\beta^2\right)^2=\left(21-8\sqrt{5}\right)^2=\left(761-336\sqrt{5}\right).$$

So, new equation whose roots are above two :

$$x^2-\left(\alpha^4+\beta^4\right)x+\left(\alpha^4\beta^4\right)=0\ .$$

or, $$x^2-\left(761+336\sqrt{5}+761-336\sqrt{5}\right)x+\left(761+336\sqrt{5}\right)\left(761-336\sqrt{5}\right)=0\ .$$

or, $$x^2-1522x+14641=0\ .$$

A is correct choice.