Find two

Find two consecutive natural numbers whose product is 1 larger than their sum. Searched numbers expressed by a fraction whose numerator is the difference between these numbers and the denominator is their sum.

Result

a =  2
b =  3

Solution:

b=a+1 ab=1+(a+b) a(a+1)=1+(a+(a+1))  a (a+1)=1+(a+(a+1)) a2a2=0  p=1;q=1;r=2 D=q24pr=1241(2)=9 D>0  a1,2=q±D2p=1±92 a1,2=1±32 a1,2=0.5±1.5 a1=2 a2=1   Factored form of the equation:  (a2)(a+1)=0  a>0 a=a1=2b=a+1 \ \\ ab=1+ (a+b) \ \\ a*(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}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 1 \pm \sqrt{ 9 } }{ 2 } \ \\ a_{1,2}=\dfrac{ 1 \pm 3 }{ 2 } \ \\ a_{1,2}=0.5 \pm 1.5 \ \\ a_{1}=2 \ \\ a_{2}=-1 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (a -2) (a +1)=0 \ \\ \ \\ a>0 \ \\ a=a_{1}=2

Checkout calculation with our calculator of quadratic equations.

b=a+1=2+1=3b=a+1=2+1=3



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