Solve x^2+8x+20=0

Question

I need to solve this problem by factoring.

Answer 1

Solve x^2 + 8x + 20 = 0 by factoring.

If the equation is factorable over the integers then
x^2 + 8x + 20 = (x + a)(x + b) = x^2 + (a+b)x + ab

ab = 20 = 2*2*5, a + b = 8

a . . . b . . . a + b
2 . . . 10 . . . 12
4 . . . 5 . . . . 9

There are no more integer factors of 20 to use, so it’s not factorable over the integers.

Since the coefficients of the equation are integers any imaginary roots will occur as complex conjugates.

x^2 + 8x + 20 = (x + a + bi)(x + a - bi) = x^2 + 2ax + a^2 + b^2

2a = 8, a = 4
a^2 + b^2 = 20, 16 + b^2 = 20, b^2 = 4, b = ±2

So x^2 + 8x + 20 = (x + 4 + 2i)(x + 4 - 2i) = 0

By Zero Product Property:

x + 4 + 2i = 0 or x + 4 - 2i = 0

x = -4 - 2i or x = -4 + 2i

Answer 2

you can use the quadratic formula but if you have to factor do the following:

x^2+8x+20=0

add -4 to both sides of the equation to complete the square

x^2+8x+20+(-4)=0+(-4)

x^2+8x+16=-4

(x+4)^2=-4

take the square root of both sides

x+4=+sqrt(-4)

x+4=+sqrt[(4)(-1)]

x+4=+2sqrt(-1) where sqrt(-1)=i, a complex number

x+4=+2i

x=-4+2i

x=-4+2i and x=-4-2i

you would get the same answer using the quadratic formula