Question

Sum of two digits is 13 and their product is 42 find nubers

Answer 1

a+b =13
a=13-b
a*b=42
(13-b)*b=42
13b-b^2=42
b^2-13b+42=0
b^2-6b - 7b+42=0
b(b-6)-7(b-6)=0
(b-6)*(b-7)=0
b=6 b=7

Answer 2

$(a + b) = 13...........(1) \\ ab = 42 \\ (a - b) {}^{2} = (a + b) {}^{2} - 4ab \\ \: \: = 13 {}^{2} - 4 \times 42$
$= 169 - 168 \\ = 1 \\ i.e.. \: (a - b) = 1.......(2) \\ \\ (1) + (2)gives \: 2a = 14 \\ a = 7 \\ and \: (1) - (2)gives \: 2b = 12 \\ b6$