Oxford University MAT 2018 Solved Papers

We have $\mathit{a}{\mathit{x}}^2+\mathit{c}=\mathit{b}{\mathit{x}}^2+\mathit{d}$ if and only if $\left(\mathit{a}-\mathit{b}\right){\mathit{x}}^2=\left(\mathit{d}-\mathit{c}\right)$. If $\left(\mathit{a}-\mathit{b}\right)=0$ then we either have no solutions (if $\mathit{c}\ne \mathit{d})$ or infinitely many solutions (if $\mathit{c}=\mathit{d}$ ). Otherwise, we can safely divide by $\left(\mathit{a}-\mathit{b}\right)$ and get ${\mathit{x}}^2=\left(\mathit{d}-\mathit{c}\right)∕\left(\mathit{a}-\mathit{b}\right)$. Then there are two roots precisely when $\left(\mathit{d}-\mathit{c}\right)∕\left(\mathit{a}-\mathit{b}\right)>0$. Multiplying up by ${\left(\mathit{a}-\mathit{b}\right)}^2$, which is positive, our expression becomes $\left(\mathit{d}-\mathit{c}\right)\left(\mathit{a}-\mathit{b}\right)>0$. The answer is (e).