Question

solve for ( x ) in the equation ( x^2 + 2x + 1 = 17 )
( \bigcirc ) ( x = -1 pm sqrt{16} )
( \bigcirc ) ( x = -1 pm sqrt{17} )
( \bigcirc ) ( x = -2 pm 2sqrt{8} )
( \bigcirc ) ( x = -1 pm sqrt{13} )

Explanation:

Step1: Rearrange the equation

$x^2 + 2x + 1 - 17 = 0$
$x^2 + 2x - 16 = 0$

Step2: Use quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here $a=1, b=2, c=-16$

Step3: Calculate discriminant

$\Delta = 2^2 - 4\times1\times(-16) = 4 + 64 = 68$

Step4: Simplify root and solve

$\sqrt{68} = \sqrt{4\times17} = 2\sqrt{17}$
$x = \frac{-2\pm2\sqrt{17}}{2} = -1\pm\sqrt{17}$

Answer:

$x=-1\pm\sqrt{17}$ (the second option)