Question

Find the TWO integers whos product is -12 and whose sum is 1
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Answer 1

Answer:

$\rm Numbers = 4 \ and \ -3.$

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 :-

$\rm\implies 1st \ number * 2nd \ number= -12\\\\\rm\implies x(1-x)=-12\\\\\rm\implies x - x^2=-12\\\\\rm\implies x^2-x-12=0\\\\\rm\implies x^2-4x+3x-12=0\\\\\rm\implies x(x-4)+3(x-4)=0\\\\\rm\implies (x-4)(x+3)=0\\\\\rm\implies\boxed{\red{\rm x = 4 , -3 }}$

Hence the numbers are 4 and -3

Answer 2

Answer:

$\displaystyle( {x}_{1} , y_{1}) =( - 3,4)\\ (x _{2}, y_{2}) = (4, - 3)$

Step-by-step explanation:

we are given two conditions

1. two integers whos product is -12
2. two integers whos sum is 1

let the two integers be x and y respectively according to the first condition

$\displaystyle xy = - 12$

according to the second condition:

$\displaystyle x + y = 1$

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:

$\displaystyle y = 1 - x$

now substitute the got value of y to the first equation which yields:

$\displaystyle x(1 - x) = - 12$

distribute:

$\displaystyle x- {x}^{2} = - 12$

add 12 in both sides:

$\displaystyle x- {x}^{2} + 12 = 0$

rearrange it to standard form:

$\displaystyle - {x}^{2} + x + 12 = 0$

divide both sides by -1:

$\displaystyle {x}^{2} - x - 12 = 0$

factor:

$\displaystyle ({x} + 3)(x - 4) = 0$

by Zero product property we acquire:

$\displaystyle {x} + 3 = 0\\ x - 4= 0$

solve the equations for x therefore,

$\displaystyle {x}_{1} = - 3\\ x _{2} = 4$

when x is -3 then y is

$\displaystyle y _{1}= 1 - ( - 3)$

simplify

$\displaystyle y _{1}= 4$

when x is 4 y is

$\displaystyle y _{2}= 1 - ( 4)$

simplify:

$\displaystyle y _{2}= - 3$

hence,

$\displaystyle( {x}_{1} , y_{1}) =( - 3,4)\\ (x _{2}, y_{2}) = (4, - 3)$