Fourth Power of Roots

Let the roots of the equation $x^2$ - cx + d = 0 be $\alpha$ and $\beta$ and the roots of the equation $x^2$ + ax + b = 0 be $\alpha_{1}$ and $\beta_{1}$ .

Then $\alpha$ = $\alpha_{1}^{4}$ and $\beta$ = $\beta_{1}^{4}$ .

Also $\alpha$ + $\beta$ = c , $\alpha \times \beta$ = d , $\alpha_{1}$ + $\beta_{1}$ = -a and $\alpha_{1} \times \beta_{1}$ = b.

Now Discriminant(D) of $x^{2}$ - 4bx + 2 $b^{2}$ -c =0 will be = $(4b)^{2}$ - $4 \times (2b^{2} - c)$

D = 16 $b^{2}$ - 8 $b^{2}$ + 4c

D = 8 $b^{2}$ + 4( $\alpha$ + $\beta$ )

D = 8 $b^{2}$ + 4( $\alpha_{1}^{4}$ + $\beta_{1}^{4}$ )

As we can see, D will be always positive. And hence, we conclude that roots of this equation are real (So the option of imaginary roots is excluded).

Now, the product of the roots of this equation is = 2 $b^2$ - c

= 2( $\alpha_{1}^{2} \times \beta_{1}^{2}$ ) - ( $\alpha$ + $\beta$ )

= 2( $\alpha_{1}^{2} \times \beta_{1}^{2}$ ) - $\alpha_{1}^{4}$ - $\beta_{1}^{4}$

= - $((\alpha_{1})^{2} + (\beta_{1})^{2})^{2}$

As this quantity is always negative, the roots must have opposite signs, i.e., one root must be positive and one root must be negative.

So, the answer is One Positive and One Negative Real Roots .

Let the roots of both quadratics be $0$ and $1$ , since it's just a special case of the problem, which shouldn't change the answer.

Therefore, $b=0$ and $c=1$ , and the final quadratic is $x^2-1=0$ . The roots of the quadratic are $x=\pm 1$ , which contains one positive and one negative real root.

As a side note, this problem wasn't worded very well, since a root could be $0$ . Maybe the choices should've been rewritten as non-negative and non-positive instead of positive and negative.