For this case we must find the solution of the following quadratic equation:
[tex]x ^ 2-10x + 25 = 0[/tex]
Where:
[tex]a = 1\\b = -10\\c = 25[/tex]
Then, the solution is given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting the values:
[tex]x = \frac {- (- 10) \pm \sqrt {(- 10) ^ 2-4 (1) (25)}} {2 (1)}\\x = \frac {10 \pm \sqrt {100-100}} {2}\\x = \frac {10 \pm \sqrt {0}} {2}\\x = \frac {10} {2} = 5[/tex]
Thus, we have two equal real roots.
[tex]x_ {1} = 5\\x_ {2} = 5[/tex]
Answer:
We have two equal real roots.
