2 Answers

Pitfall · Answer 1 · 2018-12-31T23:44:07+0000

For this case we have to:

Given the quadratic equation of the form:

ax ^ 2 + bx + c = 0

The roots are given by:

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

If we have:
x ^ 2 + 14x + 17 = -96

We can rewrite it in the following way:

$x ^ 2 + 14x + 17 + 96 = 0\\x ^ 2 + 14x + 113 = 0$

Where:

$a = 1\\b = 14\\c = 113$

Where we have:

$x = \frac {-14 \pm \sqrt {(14) ^ 2-4 (1) (113)}} {2 (1)}$

$x = \frac {-14 \pm \sqrt {(196-452)}} {2}$

$x = \frac {-14 \pm \sqrt {-256}} {2}$

By definition:
$\sqrt {-1} = i$

$x = \frac {-14 \pm \sqrt {256} i} {2}$

$x = \frac {-14 \pm16i} {2}$

$x = \frac {-14} {2} \pm \frac {16i} {2}$

$x = -7 \pm8i$

Thus, the roots are given by imaginary numbers:

$x_ {1} = - 7 + 8i\\x_ {2} = - 7-8i$

Answer:

$x_ {1} = - 7 + 8i\\x_ {2} = - 7-8i$

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