Question

Solve for x in the equation x2+2x+ 1 = 17.

Answer 1

For this case we must solve the following quadratic equation:

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

Subtracting 17 from both sides of the equation we have:

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

The solution of the equation is given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Where:

$a = 1\\b = 2\\c = -16$

Substituting the values we have:

$x = \frac {-2 \pm \sqrt {2 ^ 2-4 (1) (- 16)}} {2 (1)}\\x = \frac {-2 \pm \sqrt {4 + 64}} {2}\\x = \frac {-2 \pm \sqrt {68}} {2}\\x = \frac {-2 \pm \sqrt {2 ^ 2 * 17}} {2}\\x = \frac {-2 \pm2 \sqrt {17}} {2}\\x = -1 \pm \sqrt {17}$

Thus, we have two roots:

$x_ {1} = - 1+ \sqrt {17}\\x_ {2} = - 1- \sqrt {17}$

Answer:

$x_ {1} = - 1+ \sqrt {17}\\x_ {2} = - 1- \sqrt {17}$