Bi-quadratic Equation

The standard form of a Bi-quadratic Equation is a $x^{4}$ + b $x^{3}$ + c $x^{2}$ + dx + e = 0. Let the roots of this equation be $\alpha$ , $\beta$ , $\gamma$ and $\delta$ . Then,

The sum of its roots ( $\alpha$ + $\beta$ + $\gamma$ + $\delta$ ) = $\frac{-b}{a}$ .

The sum of its roots taken 2 at a time ( $\alpha$ $\beta$ + $\beta$ $\gamma$ + $\gamma$ $\delta$ + $\alpha$ $\gamma$ + $\alpha$ $\delta$ + $\beta$ $\delta$ ) = $\frac{c}{a}$ .

The sum of its roots taken 3 at a time ( $\alpha$ $\beta$ $\gamma$ + $\alpha$ $\gamma$ $\delta$ + $\alpha$ $\beta$ $\delta$ + $\beta$ $\gamma$ $\delta$ ) = $\frac{-d}{a}$ .

And the product of its roots ( $\alpha$ $\beta$ $\gamma$ $\delta$ ) = $\frac{e}{a}$ .

In the given equation, a = 1, b = -4, c = 8, d= 9 and e = 11. So,

$\alpha$ + $\beta$ + $\gamma$ + $\delta$ = 4.

$\alpha$ $\beta$ + $\beta$ $\gamma$ + $\gamma$ $\delta$ + $\alpha$ $\gamma$ + $\alpha$ $\delta$ + $\beta$ $\delta$ = 8.

$\alpha$ $\beta$ $\gamma$ + $\alpha$ $\gamma$ $\delta$ + $\alpha$ $\beta$ $\delta$ + $\beta$ $\gamma$ $\delta$ = -9.

$\alpha$ $\beta$ $\gamma$ $\delta$ = 11.

Now, $\alpha^{2}$ + $\beta^{2}$ + $\gamma^{2}$ + $\delta^{2}$ = { $\alpha$ + $\beta$ + $\gamma$ + $\delta$ }^2 - 2( $\alpha$ $\beta$ + $\beta$ $\gamma$ + $\gamma$ $\delta$ + $\alpha$ $\gamma$ + $\alpha$ $\delta$ + $\beta$ $\delta$ ).

= $4^{2}$ - 2(8)

= 16 - 16 = 0

So, $\alpha^{2}$ + $\beta^{2}$ + $\gamma^{2}$ + $\delta^{2}$ + $\alpha$ $\beta$ $\gamma$ $\delta$ = 0 + 11 = 11

And we have got our answer ........ $\boxed{11}$