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.
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.
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.`