Question

Find the solution set of the quadratic equation over the set of complex numbers. 5x2 + 8x + 4 = 0 A) x = −1 − i 2 or −1 + i 2 B) x = − 4 5 − 2i 5 or − 4 5 + 2i 5 C) x = − 2 5 − 6i 5 or − 2 5 + 6i 5 D) x = − 1 4 − 3i 4 or − 1 4 + 3i 4

Answer 1

5x^2 + 8x + 4 = 0 is a quadratic. a=5, b=8 and c=4

Thus,

-8 plus or minus sqrt( 8^2-4(5)(4) )
x = ------------------------------------------------
2(5)
-8 plus or minus sqrt(64-80)
= ---------------------------------------------
10


-8 plus or minus sqrt(-16) -8 plus or minus i*4
= -------------------------------------- = -----------------------------
10 10
-4 plus or minus i*2
= ------------------------------ Can you now choose the correct answer?
5

Answer 2

The solutions to the quadratic equation are:

`x=-(4)/(5)+i(2)/(5),\:x=-(4)/(5)-i(2)/(5)`

Explanation:

Complex numbers are numbers of the form
`a+bi`, where
`a` and
`b` are real numbers.

For a quadratic equation of the form
`ax^2+bx+c=0` the solutions are

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

If
`b^2-4ac<0`, the equation has two complex solutions that are not real.

Quadratic equations with a negative discriminant have no real number solution. However, if we extend our number system to allow complex numbers, quadratic equations will always have a solution.

This quadratic equation
`5x^2\:+\:8x\:+\:4\:=\:0` is not factorable, so we apply the quadratic formula.

`\mathrm{For\:}\quad a=5,\:b=8,\:c=4\\\\x_(1,\:2)=(-8\pm √(8^2-4\cdot \:5\cdot \:4))/(2\cdot \:5)\\`

`x_1=(-8+√(8^2-4\cdot \:5\cdot \:4))/(2\cdot \:5)=(-8+√(16)i)/(2\cdot \:5)=-(4)/(5)+(2)/(5)i`

`x_2=(-8-√(8^2-4\cdot \:5\cdot \:4))/(2\cdot \:5)=(-8-√(16)i)/(10)=-(4)/(5)-(2)/(5)i`

The solutions to the quadratic equation are

`x=-(4)/(5)+i(2)/(5),\:x=-(4)/(5)-i(2)/(5)`