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Given the quadratic equation:

x^2 - 10x + 25 = 0

We notice that the equation is a complete square such that:

x^2 - 10x + 25 = (x-5)^2

==> (x-5)^2 = 0

==> x = 5

**Then the equation has only one roots which is x= 5.**

To find the roots for x^2-10x+25:

Method I: Recognize that this is a perfect square trinomial of the form a^2 - 2ab + b^2, which factors as (a-b)(a-b) with a=x and b=5.

Thus the solution to x^2-10x+25=0 is the solution to (x-5)^2=0 which is 5. The root is the solution(s) of the equation.

Methosd II: We need two numbers whose product is 25 and whose sum is -10. The numbers are -5 and -5 since (-5)(-5)=25 and -5+(-5)=-10. Then the given trinomial factors as (x-5)(x-5) and the solution is obtained as above.

You could also apply quadratic formula to determine the factors of the given quadratic.

Let's recall the formula first:

`x_(1,2) = [-b+-sqrt(b^2 -4ac )]/(2a)`

We'll identify the coefficients a,b,c:

a = 1 ; b = -10 and c = 25

`x_(1,2) = (10+-sqrt(100 - 100))/2`

`x_(1,2) = 10/2`

`x_1 = x_2 = 5`

We notice that the roots of equation are equal, therefore we can re-write the quadratic:

`(x - x_1)(x - x_2) = 0`

`(x - 5)(x - 5) = 0`

`(x - 5)^2 = 0`

**Therefore, the quadratic equation will have two equal solutions `x_1 = x_2 = 5.` **

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