Two numbers with given difference and product

Text

Find two numbers whose difference is $1$ and whose product is $13$ (note: $a-b$ can be written as $a+(-b)$).

Solution

Call the two numbers $a$ and $b$, with $a-b=1$ and $ab=13$. The suggested trick is to consider the pair $\{a,\,-b\}$: for it $\text{sum}=a+(-b)=a-b=1,\qquad \text{product}=a\cdot(-b)=-ab=-13.$ Hence $a$ and $-b$ are the roots (by Vieta’s formulas) of $t^2-1\cdot t+(-13)=0 \;\Longrightarrow\; t^2-t-13=0.$ Discriminant: $\Delta=(-1)^2-4\cdot1\cdot(-13)=1+52=53>0,$ so $t=\frac{1\pm\sqrt{53}}{2}.$ Numerically, $\sqrt{53}\approx7.280$, giving $t_1\approx4.140$ and $t_2\approx-3.140$. The two roots are $a$ and $-b$: taking $a=t_1\approx4.140$ gives $-b=t_2\approx-3.140$, that is $b\approx3.140$. Check: difference $a-b\approx4.140-3.140=1$; product $a\,b\approx4.140\cdot3.140\approx13$. ✓ $\boxed{n_1=a\approx4.140,\qquad n_2=b\approx3.140}$