Question

The sum of two integer is -11 and their product is -80 what are the two integers

Answer 1

Given :

• The sum of two integer is -11
• Product → - 80

To Find :

• The two integers.

Solution :

Let the two integers be x and y.

Case 1 :

Sum of greater integer, x and smaller integer y is - 11.

Equation :

$\longrightarrow$ $\sf{x+y=-11}$

$\sf{x=-y-11\:\:(i)}$

Case 2 :

Product of x and y is - 80

Equation :

$\sf{xy=-80}$

From, equation (i)

$\longrightarrow$ $\sf{(-y-11)y=-80}$

$\longrightarrow$ $\sf{-y^2-11y=-80}$

$\longrightarrow$ $\sf{-y^2-11y+80=0}$

Divide, throughout by minus,

$\longrightarrow$ $\sf{y^2+11y-80=0}$

Now, we see that the above equation is a quadratic equation with variable y.

Solve using factorization method.

$\longrightarrow$ $\sf{y^2\:+\:16y-5y-80=0}$

$\longrightarrow$ $\sf{y(y+16)-5(y+16)=0}$

$\longrightarrow$ $\sf{(y+16)\:\:(y-5) =0}$

$\longrightarrow$ $\sf{y+16=0\:\:or\:\:y-5=0}$

$\longrightarrow$ $\sf{y=-16\:\:or\:\:y=5}$

So, now we have two values of y.

Let's discuss both cases.

When, y = - 16 :

Substitute, y =- 16 in equation (i)

$\longrightarrow$ $\sf{x=-y-11}$

$\longrightarrow$ $\sf{x=-(-16)-11}$

$\longrightarrow$ $\sf{x=16-11}$

$\longrightarrow$ $\sf{x=5}$

•°• When, y = - 16,

• x → 5
• y → - 16

When, y = 5 :

Substitute, y = 5 in equation (i),

$\longrightarrow$ $\sf{x=-y-11}$

$\longrightarrow$ $\sf{x=-5-11}$

$\longrightarrow$ $\sf{x=-16}$

•°• When, y = 5,

• x → -16
• y = 5

Answer 2

Let the 2 integers be x and y.

Given

Sum of 2 integers = -11

$=> x + y = -11$

Product of 2 integers = -80

$=> xy = -80$

Now,

$x + y = -11\\=> x = -11-y$

Substituting and solving we get,

$(-11-y)y = -80\\-y^{2} -11y+80=0\\y^{2} + 11y-80=0\\y^{2} +16y-5y-80=0\\y(y+16) - 5(y+16)=0\\(y-5)(y+16)=0$

Now,

$y - 5 = 0\\=> y = 5\\=> x = -16\\\boxed{\boxed{\bold{Therefore \ the \ 2 \ integers \ are \ 5 \ and -16}}}$