Mathematics - Quadratic Equation Question with Solution | TestHub

Since, $a+b=-p,ab=1$ .....(i) and $c+d=-q,cd=1$ Now $(a-c)(b-c)$ and $(a+d)(b+d)$ are the roots of $x^2+ax+\beta =0$ $$\begin{array}{rl}& (a-c)(b-c)(a+d)(b+d)=\beta \\ & \Rightarrow \left(ab-ac-bc+c^2\right)\left(ab+ad+bd+d^2\right)=\beta \\ & \Rightarrow \left\{1-c(a+b)+c^2\right\}\left\{1+d(a+b)+d^2\right\}=\beta \\ & \Rightarrow \left(1+pc+c^2\right)\left(1-pd+d^2\right)=\beta \\ & \Rightarrow 1-pd+d^2+pc-p^{2}cd+pc{d}^2+c^2-p{c}^{2}d\\ & +c^2{d}^2=\beta \\ & \Rightarrow 1-pd+d^2+pc-p^2+pd+c^2-pc+1=\beta \\ & [\because cd=1]\\ & \Rightarrow 2+d^2+c^2-p^2=\beta \\ & \end{array}$$ $\begin{array}{rl}\Rightarrow 2cd+c^2+d^2-p^2& =\beta \\ (c+d{)}^2-p^2& =\beta \\ q^2-p^2& =\beta [\because (c+d)=-q]\end{array}$