Find two

Find two consecutive natural numbers whose product is one larger than their sum. Searched numbers are expressed by a fraction whose numerator is the difference between these numbers, and the denominator is their sum.

Correct answer:

a =  2
b =  3

Step-by-step explanation:

$$\begin{gathered} b=a+1 \\ ab=1+(a+b) \\ a\cdot (a+1)=1+(a+(a+1)) \\ a\cdot (a+1)=1+(a+(a+1)) \\ {a}^2-a-2=0 \\ p=1;q=-1;r=-2 \\ D=q^2-4pr=1^2-4\cdot 1\cdot (-2)=9 \\ D>0 \\ {a}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{1\pm \sqrt{9}}{2} \\ {a}_{1,2}=\frac{1\pm 3}{2} \\ {a}_{1,2}=0.5\pm 1.5 \\ {a}_1=2 \\ {a}_2=-1 \\ a>0 \\ a=a_1=2 \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} b=a+1=2+1=3 \\ \text{ Verifying Solution: } \\ {s}_1=a\cdot b=2\cdot 3=6 \\ {s}_2=a+b=2+3=5 \\ {s}_1=1+s_2 \end{gathered}$$

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