Question

Solve for x in the equation x squared + 2 x + 1 = 17.

x = negative 1 plus-or-minus StartRoot 15 EndRoot
x = negative 1 plus-or-minus StartRoot 17 EndRoot
x = negative 2 plus-or-minus 2 StartRoot 5 EndRoot
x = negative 1 plus-or-minus StartRoot 13 EndRoot

Answer 1

The answer to your question is the second option

Explanation:

Process

1.- Write the equation

x² + 2x + 1 = 17

Factor the first term

(x + 1)² = 17

Get the square root

`\sqrt{(x+1)^(2) }  = √(17)`

(x + 1) =
`√(17)`

Result

x₁ = - 1 +
`√(17)`

x₂ = -1 -
`√(17)`

Answer 2

Option B.

Explanation:

The given equation is

`x^2+2x+1=17`

Subtract both sides by 17.

`x^2+2x+1-17=17-17`

`x^2+2x-16=0` .... (1)

If a quadratic equation is
`ax^2+bx+c=0`, then by quadratic formula

`x=(-b\pm √(b^2-4ac))/(2a)`

In equation (1), a=1, b=2 and c=-16. Using quadratic formula we get

`x=(-(2)\pm √((2)^2-4(1)(-16)))/(2(1))`

`x=(-2\pm √(4+64))/(2)`

`x=(-2\pm √(68))/(2)`

`x=(-2\pm 2√(17))/(2)`

Taking out common factors.

`x=(2(-1\pm √(17)))/(2)`

`x=-1\pm √(17)`

Therefore, the correct option is B.