2007-04-27 19:34:25 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2-10x+25=0
(x-5)(x-5)=0
You seem to have posted a lot of factoring questions. Please email me or im me if you want help on general factoring instead of just specific examples.
2007-04-27 19:37:47 · answer #1 · answered by Anonymous · 0⤊ 0⤋
This can be solved by completing the square:
x^2 + 25 = 10x
x^2 - 10x + 25 = 0
(x)^2 - (2*5*x) + (5)^2 = 0
(x - 5)^2 = 0
(x - 5)(x - 5) = 0
Either x - 5 = 0 (or) x - 5 = 0
x = 5 (or) x = 5
x = 5, 5
{5, 5} is the solution
x = 5 is the solution of this equation (Perfect square quadratic equations have only one solution, but the solution must be repeated in the braces)
2007-04-28 10:32:11 · answer #2 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
Sorry, mate. You got this one wrong.
x^2 + 25 = 10 x
x^2 - 10 x + 25 = 0
(x - 5)^2 = 0
x= 5
There is only one solution to this, i.e. 5
2007-04-28 02:38:25 · answer #3 · answered by Daniel T 2 · 0⤊ 0⤋
x^2 -10x + 25 = (x - 5)^2, so the answer is (5.5)
2007-04-28 02:38:09 · answer #4 · answered by gp4rts 7 · 0⤊ 0⤋
(x-5)(x-5)=0
So the answer is (5,5)
you want to find the two # that add together = -10 (-5-5=-10), and multiply together to get +25 (-5)(-5)=+25
2007-04-28 02:47:38 · answer #5 · answered by luumomo 1 · 0⤊ 0⤋
x is equal to 5 when y is 0.
2007-04-28 02:50:53 · answer #6 · answered by Anonymous · 0⤊ 0⤋
x^2-10x+25=0
x^2-5x-5x+25=0
x(x-5)-5(x-5)=0
(x-5)(x-5)=(x-5)^2=0
x-5=0
x=5
There is only one root for this quadratic equation.
2007-04-28 03:11:58 · answer #7 · answered by Harsh 2 · 0⤊ 0⤋
answer 5
2007-04-28 04:31:15 · answer #8 · answered by wrpoop 2 · 0⤊ 0⤋
(x-5)^2=0
x=5,5
2007-04-28 02:38:00 · answer #9 · answered by raj 7 · 0⤊ 0⤋
i never even thought to put my own homework questions on here... good thinking!
2007-04-28 02:43:54 · answer #10 · answered by Anonymous · 0⤊ 0⤋