Question

What two numbers multiply to 44 and add up to 12?

Answer 1

`xy=44\\ x+y=12\\\\ xy=44\\ x=12-y\\\\ (12-y)y=44\\ 12y-y^2=44\\ y^2-12y+44=0\\ y^2-12y+36+8=0\\ (y-6)^2=-8`

No solutions in real numbers.

In complex numbers:

`(y-6)^2=-8\\ y-6=-√(-8) \vee y-6=√(-8)\\ y=6-2\sqrt2 i \vee y=6+2\sqrt2 i\\\\ x=12-(6-2\sqrt2i) \vee x=12-(6+2\sqrt2i)\\ x=12-6+2\sqrt2i \vee x=12-6-2\sqrt2i\\ x=6+2\sqrt2i \vee x=6-2\sqrt2i`

These numbers are
`6-2\sqrt2i` and
`6+2\sqrt2i`.

Answer 2

This pattern of question is always coming up. Since we can't easily guess, then let us set up simultaneous equation for the statements.

let the two numbers be x and y.

Multiply to 44. x*y = 44 ..........(a)

Add up to 12. x + y = 12 .........(b)

From (b)

y = 12 - x .......(c)

Substitute (c) into (a)

x*y = 44

x*(12 - x) = 44

12x - x² = 44

-x² + 12x = 44

-x² + 12x - 44 = 0.

Multiply both sides by -1

-1(-x² + 12x - 44) = -1*0

x² - 12x + 44 = 0.

This does not look factorizable, so let us just use quadratic formula

comparing to ax² + bx + c = 0, x² - 12x + 44 = 0, a = 1, b = -12, c = 44

x = (-b + √(b² - 4ac)) /2a or (-b - √(b² - 4ac)) /2a


x = (-(-12) + √((-12)² - 4*1*44) )/ (2*1)

x = (12 + √(144 - 176) )/ 2

x = (12 + √-32 )/ 2

√-32 = √(-1 *32) = √-1 * √32 = i * √(16 *2) = i*√16 *√2 = i*4*√2 = 4i√2

Where i is a complex number. Note the equation has two values. We shall include the second, that has negative sign before the square root.

x = (12 + √-32 )/ 2 or (12 - √-32 )/ 2

x = (12 + 4i√2 )/ 2 (12 - 4i√2 )/ 2

x = 12/2 + (4i√2)/2 12/2 - (4i√2)/2

x = 6 + 2i√2 or 6 - 2i√2

Recall equation (c):

y = 12 - x, When x = 6 + 2i√2, y = 12 - (6 + 2i√2) = 12 - 6 - 2i√2 = 6 - 2i√2

When x = 6 - 2i√2, y = 12 - (6 - 2i√2) = 12 - 6 + 2i√2 = 6 + 2i√2


x = 6 + 2i√2, y = 6 - 2i√2

x = 6 - 2i√2, y = 6 + 2i√2

Therefore the two numbers that multiply to 44 and add up to 12 are:

6 + 2i√2 and 6 - 2i√2