Question

PLEASE HELP ITS URGENT

Solve for the roots in simplest form using the quadratic formula:

x²-16x = -100

Answer 1

`\rm x = 8 \pm 6i`

Explanation:

Given equation: x² - 16x = -100

x² - 16x + 100 = 0

By comparing given equation with standard quadratic equation i.e. ax² + bx + c = 0 we get:

a = 1

b = -16

c = 100

Quadratic formula:

`\boxed{ \rm x = \frac{ - b \pm \sqrt{ {b}^(2) - 4ac } }{2a} }`

Now, substituting the values in the quadratic formula:

`\rm \implies x = \frac{ - ( - 16) \pm \sqrt{ {( - 16)}^(2) - 4(1)( 100) } }{2(1)} \\ \\ \rm \implies x = ( 16 \pm √( 256 - 400) )/(2) \\ \\ \rm \implies x = ( 16 \pm √(-144 ) )/(2) \\ \\ \rm \implies x = ( 16 \pm 12i )/(2) \\ \\ \rm \implies x = 8 \pm 6i`

Answer 2

x has no real solutions.

x = 8 + 6i, x = 8 - 6i

Explanation:

First, move the all the terms to one side of the equation.

x^2 - 16x + 100 = -100 + 100

x^2 - 16x + 100 = 0

Then, based on our knowledge of the standard form of a quadratic equation: ax^2 + by + c = 0, we can plug the coefficients in front of the variables into the formula, which looks like
`x=(-b+or-√(b^2-4ac) )/(2a)`.

Our a here is 1,

The b is -16,

The c is 100.

plugging it in:

`(-(-16)+or-√((-16)^2-4*1*100) )/(2*1)`

simplifies down to:

16/2 + or - (√(256 - 400))/2

= 8 + or - √(-36)

Here, we have a negative square root, meaning there will be no roots for this equation in the real number system.

If you include imaginary/complex numbers, this equation will have roots.

x = 8 + or - √(-36)

x = 8 + or - 6√(-1)

x = 8 + or - 6i

so the final answer:

x = 8 + 6i, x = 8 - 6i