Answer:
[tex] \rm Numbers = 4 \ and \ -3.[/tex]
Step-by-step explanation:
Given :-
The sum of two numbers is 1 .
The product of the nos . is 12 .
And we need to find out the numbers. So let us take ,
First number be x
Second number be 1-x .
According to first condition :-
[tex]\rm\implies 1st \ number * 2nd \ number= -12\\\\\rm\implies x(1-x)=-12\\\\\rm\implies x - x^2=-12\\\\\rm\implies x^2-x-12=0\\\\\rm\implies x^2-4x+3x-12=0\\\\\rm\implies x(x-4)+3(x-4)=0\\\\\rm\implies (x-4)(x+3)=0\\\\\rm\implies\boxed{\red{\rm x = 4 , -3 }}[/tex]
Hence the numbers are 4 and -3
Answer:
[tex] \displaystyle( {x}_{1} , y_{1}) =( - 3,4)\\ (x _{2}, y_{2}) = (4, - 3)[/tex]
Step-by-step explanation:
we are given two conditions
let the two integers be x and y respectively according to the first condition
[tex] \displaystyle xy = - 12[/tex]
according to the second condition:
[tex] \displaystyle x + y = 1[/tex]
now notice that we have two variables therefore ended up with a simultaneous equation so to solve the simultaneous equation cancel x from both sides of the second equation which yields:
[tex] \displaystyle y = 1 - x[/tex]
now substitute the got value of y to the first equation which yields:
[tex] \displaystyle x(1 - x) = - 12[/tex]
distribute:
[tex] \displaystyle x- {x}^{2} = - 12[/tex]
add 12 in both sides:
[tex] \displaystyle x- {x}^{2} + 12 = 0[/tex]
rearrange it to standard form:
[tex] \displaystyle - {x}^{2} + x + 12 = 0[/tex]
divide both sides by -1:
[tex] \displaystyle {x}^{2} - x - 12 = 0[/tex]
factor:
[tex] \displaystyle ({x} + 3)(x - 4) = 0[/tex]
by Zero product property we acquire:
[tex] \displaystyle {x} + 3 = 0\\ x - 4= 0[/tex]
solve the equations for x therefore,
[tex] \displaystyle {x}_{1} = - 3\\ x _{2} = 4[/tex]
when x is -3 then y is
[tex] \displaystyle y _{1}= 1 - ( - 3)[/tex]
simplify
[tex] \displaystyle y _{1}= 4[/tex]
when x is 4 y is
[tex] \displaystyle y _{2}= 1 - ( 4)[/tex]
simplify:
[tex] \displaystyle y _{2}= - 3[/tex]
hence,
[tex] \displaystyle( {x}_{1} , y_{1}) =( - 3,4)\\ (x _{2}, y_{2}) = (4, - 3)[/tex]
