If the two roots of the equation ${{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ are such that one is the negative of the other, show that ${{c}^{2}}=a(bc-ad)$
$\begin{align} & \text{Let the roots be (}x-p)\text{ and }(x+p)\,\text{of the equation} \\ & \text{When }(x-p)\text{ is a factor} \\ & f(p)={{p}^{4}}+a{{p}^{3}}+b{{p}^{2}}+cp+d=0---(i) \\ & \text{When }(x+p)\text{ is a factor} \\ & f(-p)={{p}^{4}}-a{{p}^{3}}+b{{p}^{2}}-cp+d=0---(ii) \\ & \text{Add }(i)\text{ and }(ii) \\ & 2{{p}^{4}}+2b{{p}^{2}}+2d=0 \\ & {{p}^{4}}+b{{p}^{2}}+d=0-----(iii) \\ & \text{Subtract }(ii)\text{ from }(i) \\ & 2a{{p}^{3}}+2cp=0 \\ & 2p(a{{p}^{2}}+c)=0----(iv) \\ & p=0\text{ or }a{{p}^{2}}=-c \\ & {{p}^{2}}=-\frac{c}{a}----(v) \\ & \text{Substitute }{{p}^{2}}=-\frac{c}{a}\text{ into (iii)} \\ & {{\left( -\frac{c}{a} \right)}^{2}}+b\left( -\frac{c}{a} \right)+d=0 \\ & \frac{{{c}^{2}}}{{{a}^{2}}}-\frac{bc}{a}+d=0 \\ & {{c}^{2}}-abc+{{a}^{2}}d=0 \\ & {{c}^{2}}=abc-{{a}^{2}}d \\ & {{c}^{2}}=a(bc-ad) \\\end{align}$