# What two integers whose sum is -21 and whose product is -4?

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### 1 Answer

Let x and y be the two integers.

Then, let's represent the given conditions in math form.

The first condition is "sum of two integers is -21". So,

`x+y=-21` (Let this be EQ1.)

And the second condition is "product of the two integers is -4". In math form, it is:

`xy=-4` (Let this be EQ2.)

Then, let's use the substitution method of system of equations. From EQ2, substitute `y=-4/x` to EQ1.

`x+y=-21`

`x+(-4/x)=-21`

`x-4/x=-21`

Multiply both side by x to cancel the denominator.

`x(x-4/x)=-21x`

`x^2-4=-21x`

Express the equation in quadratic form `ax^2+bx+c=0` .

`x^2+21x-4=0`

Use quadratic formula to solve for x.

`x=[-b+-sqrt(b^2-4ac)]/(2a)= [-21+-sqrt(21^2-4(1)(-4))]/(2*1) = [-21+sqrt(441+16)]/2 =(-21+-sqrt457)/2`

`x_1= (-21 + sqrt457)/2 ` and `x_2=(-21 - sqrt457)/2`

Substitute values of x to y=-4/x

`x_1=(-21+sqrt457)/2` , `y_1=-4/((-21+sqrt457)/2) = -8/(-21+sqrt457)`

To simplify y1, multiply top and bottom by the conjugate of the denominator.

`y_1= -8/(-21+sqrt457)*(-21-sqrt457)/(-21-sqrt457)`

`y_1=-(-168-8sqrt457)/(-16) = -(21+sqrt457)/2`

`y_1=(-21-sqrt457)/2`

`x_2=(-21-sqrt457)/2` , `y_2=-4/(-21-sqrt457/2) = 8/(21+sqrt457)`

Multiply the top and bottom of y2 by the conjugate of the denominator too.

`y_2 = 8/(21+sqrt457)*(21-sqrt457)/(21-sqrt457)`

`y_2= (168 - 8sqrt457)/ (-16) = (-21 +sqrt457)/2`

**Hence the possible values of the two integers are:**

**`x_1=(-21+sqrt457)/2` , `y_1=(-21-sqrt457)/2` and**

**`x_2=(-21-sqrt457)/2` , `y_2=(-21+sqrt457)/2` **