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jeyben [28]
3 years ago
9

2. y = k2 - 13k +42​

Mathematics
1 answer:
bekas [8.4K]3 years ago
6 0

For this case we must solve the following quadratic equation:

y = k ^ 2-13k + 42

With y = 0 we have:

k ^ 2-13k + 42 = 0

The roots will be given by:

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

Where:

a = 1\\b = -13\\c = 42

Substituting:

k = \frac {- (- 13) \pm \sqrt {(- 13) ^ 2-4 (1) (42)}} {2 (1)}\\k = \frac {13 \pm \sqrt {169-168}} {2}\\k = \frac {13 \pm \sqrt {1}} {2}\\k = \frac {13 \pm1} {2}

Thus, we have two roots:

k_ {1} = \frac {13 + 1} {2} = \frac {14} {2} = 7\\k_ {2} = \frac {13-1} {2} = \frac {12} {2} = 6

Answer:

k_ {1} = 7\\k_ {2} = 6

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