Question

2. y = k2 - 13k +42​

Answer 1

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`