For this case we must solve the following quadratic equation:
[tex]y = k ^ 2-13k + 42[/tex]
With [tex]y = 0[/tex] we have:
[tex]k ^ 2-13k + 42 = 0[/tex]
The roots will be given by:
[tex]k = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Where:
[tex]a = 1\\b = -13\\c = 42[/tex]
Substituting:
[tex]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}[/tex]
Thus, we have two roots:
[tex]k_ {1} = \frac {13 + 1} {2} = \frac {14} {2} = 7\\k_ {2} = \frac {13-1} {2} = \frac {12} {2} = 6[/tex]
Answer:
[tex]k_ {1} = 7\\k_ {2} = 6[/tex]
