Question

1.)What number needs to be added to both sides of the equation in order to complete the square? x2+16x=18

answer is 64

x^2+16x+64=18+64

2.)Solve for x over the complex numbers.

x2+10x+41=0

answer is x=-5+4i and -5-4i

3.)What is the factored form of the expression over the complex numbers?

16x2+9y2
answer is (4x+3iy)(4x-3iy)

Answer 1

all of your answers are correct

1.) 64

2.) x= -5+4i and x= -5-4i

3.) (4x+3iy)(4x-3iy)

Answer 2

1.

When we are completing squares, we need to divide by 2 the linear term and then find its square power, that's the term we need to add on both sides of the equality, as follows

`((16)/(2))^(2) =(8)^(2)=64`

Basically, we need to add the number 64 both sides

`x^(2) +16x+64=18+64`

2.

The given equation is

`x^(2) +10x+41=0`

We need to apply the quadratic formula to solve this equation

`x_(1,2) =\frac{-b(+-)\sqrt{b^(2)-4ac } }{2a}`

Where
`a=1`,
`b=10` and
`c=41`. Replacing these values, we have

`x_(1,2) =\frac{-10(+-)\sqrt{10^(2)-4(1)(41) } }{2(1)}\\x_(1,2) =(-10(+-)√(100-164 ) )/(2)=(-10(+-)√(-64) )/(2)`

There we need to use complex number, to transform the subradical number in a positive number

`x_(1,2)=(-10(+-)√(64)i )/(2)=(-10(+-)8i)/(2)\\ x_(1,2)=-5(+-)4i`

Therefore, the complex solutions are

`x_(1)=-5+4i\\ x_(1)=-5-4i`

3.

The given expression is

`16x^(2) +9y^(2)`

To solve this expression, remember that
`i=√(-1)`

First, we expresse both squares uniformly,

`16x^(2) +9y^(2)=(4x)^(2)+(3y)^(2)`

But, we know that
`-(-1)=1`, so

`(4x)^(2)+(3y)^(2)=(4x)^(2)-(-1)(3y)^(2)`

Then,

`(4x)^(2)-(-1)(3y)^(2)=(4x)^(2)-(3y)^(2)i^(2)`, because
`i^(2)=-1`

Therefore, the expression with complex numbers is

`(4x)^(2)-(3iy)^(2)\\\therefore (4x+3iy)(4x-3iy)`