Question

Solve for x in the equation x^2+2x+1=17

Answer 1

First, subtract 17 on both sides: x²+2x-16=0
this cannot be factored, so use the quadratic formula to solve for x:
b²-4ac=2²-4(10(-16)=4+64=68
√68=2√17
so x=(-2+2√17)/2 or x=(-2-2√17)/2
x=-1+√17 or x=-1-√17

Answer 2

Answer:

 $x=-1+\sqrt{17}$   and   $x=-1-\sqrt{17}$

Step-by-step explanation:

$x^2+2x+1=17$

To solve for x , we set the right hand side =0

$x^2+2x+1=17$

Subtract 17 on both sides

$x^2+2x-16=0$

Now apply quadratic formula

$x=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

From the given f(x), the value of a=1, b=2, c=-16

Plug in all the values in the formula

$x=\frac{-2+-\sqrt{2^2-4(1)(-16)}}{2(1)}$

 $x=\frac{-2+-\sqrt{68}}{2}$

 $x=\frac{-2+-2\sqrt{17}}{2}$

 $x=-1+-\sqrt{17}$

 $x=-1+\sqrt{17}$   and   $x=-1-\sqrt{17}$