2 years ago

Will upvote!

Mathematics

2 answers:

My name is Ann [436]2 years ago

7 0

(-2,8),(-1,5),(0,2),(1,-1),(2,-4) are points of the line.

Natasha_Volkova [10]2 years ago

6 0

−7y-21x=−6-8
−7y-21x =−14

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Solve the equation: k^2+5k+13=0

mr_godi [17]

Step-by-step explanation:

k² + 5k + 13 = 0

Using the quadratic formula which is

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

From the question

a = 1 , b = 5 , c = 13

So we have

$k = \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)} \\ = \frac{ - 5 \pm \sqrt{25 - 52} }{2} \\ = \frac{ - 5 \pm \sqrt{ - 27} }{2} \: \: \: \: \: \: \\ = \frac{ - 5 \pm3 \sqrt{3} \: i}{2} \: \: \: \: \: \:$

Separate the solutions

$k_1 = \frac{ - 5 + 3 \sqrt{3} \: i }{2} \: \: \: \: or \\ k_2 = \frac{ - 5 - 3 \sqrt{3} \: i}{2}$

The equation has complex roots

Separate the real and imaginary parts

We have the final answer as

$k_1 = - \frac{5}{2} + \frac{3 \sqrt{3} }{2} \: i \: \: \: \: or \\ k_2 = - \frac{5}{2} - \frac{3 \sqrt{3} }{2} \: i$

Hope this helps you

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