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Marina86 [1]
3 years ago
13

Solve x2 + 14x + 17 = –96 for x

Mathematics
2 answers:
jarptica [38.1K]3 years ago
5 0

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


sergiy2304 [10]3 years ago
4 0
X=-7±4 Square root of 6i
x = –7 ± 8i
x= 7± 4 Square root of 6i
<span>x = 7 ± 8i</span>
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