Question

Find two integers whose sum is 7 and product is 12

Answer 1

Let the 2 integers be x and y.

Given

Sum of 2 integers = 7

$\implies x + y = 7$

Product of 2 integers = 12

$\implies xy=12$

Solving, we get,

$y = 7-x\\\implies x(7-x) = 12\\\implies -x^{2} +7x-12=0\\\implies x^{2} -7x+12=0\\\implies x^{2} -4x-3x+12=0\\\implies x(x-4)-3(x-4)=0\\\implies (x-3)(x-4)=0$

When x = 3, y = 4

When x = 4, y = 3

                                                       

Answer 2

↪Question :-

• Find two integers whose sum is 7 and product is 12 .

↪Solution :-

Given :

• Sum = 7
• Product = 12

To Find :

• The two integers.

Let the two integers be x and y.

$\tt{Sum\:of\:the\:two\:integers=7} \\ \tt{\longrightarrow{x + y} = 7 }\\ \tt {\longrightarrow{y = 7 - x \: \: \: \: \: \: \: ...(i)}}$

$\tt{Product \: of \: the \: two \: integer = 12} \\ \tt{ \longrightarrow {xy = 12}}$

By Putting Value of equation (i) :

$\sf{ \implies{x \times y = 12}} \\ \\ \sf{ \implies{x \times (7 - x)= 12}} \\ \\ \sf{ \implies{7x - {x}^{2} = 12 }} \\ \\ \sf{ \implies{{- x}^{2} + 7x - 12=0}} \\ \\ \sf{ \implies{{x}^{2} -7x +12=0}} \\ \\ \sf{ \implies{{x}^{2}-4x-3x+12=0}}\\ \\\sf{ \implies{x (x-4)-3 (x-4)=0}}\\ \\\sf{ \implies{(x-4)(x-3)=0}}\\ \\ \sf{ \implies{x=4 \: or\: x=3}}$

• When x = 4 , other integer = 7-4 =3
• When x = 3 , other integer = 7-3=4

$\bf{\therefore {The\:integers\:are\:3 \:and\:4.}}$

$\rule {307}{2}$