Answer:
Option B is correct
The number of real number solutions for the given equation is, 1.
Step-by-step explanation:
Given the equation: [tex]x^2-10x+25 =0[/tex]
Since, this is a quadratic equation of the form of [tex]ax^2+bx+c =0[/tex]
where a =1 , b = -10 and c =25.
The discriminant of a quadratic equation is, [tex]{b^2-4ac}[/tex]
then;
Discriminant = [tex]{(-10)^2-4(1)(25)}[/tex] = (100-100) = 0
Since, a discriminant of zero means there is only one real solution for x.
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Substitute the given values.
[tex]x = \frac{-(-10) \pm 0}{2(1)}[/tex]
[tex]x = \frac{10}{2}[/tex]
Simplify:
x = 5
Therefore, the number of real solution for the given equation is, 1
Answer:
There is one solution
Step-by-step explanation:
x2 – 10x + 25 = 0
We can factor
(x-5) (x-5) = 0
Using the zero product property
x-5=0 x-5=0
x=5 x=5
There is one solution
