Assuming the expression you want to solve is: x^2 + 10x + 25 = 81 Subtract both sides by 81: x^2 + 10x - 56=0 First, check for factors of 56 whose difference is 10: 2 and 28, no, 4 and 14, yes. So factor the expression as: (x+14)(x-4)=0 Thus the solutions to the expression are either x=4 or x=-14.
When the given expression equals 0 then x = -1/6 and x = -6
That equation can not be factored. It can however be solved for x: x2 - 9x + 4 = 0 x2 - 9x + 81/4 = 81/4 - 4 (x - 9/2)2 = 81/4 - 16/4 (x - 9/2)2 = 65/4 x - 9/2 = ± √65 / 2 x = (9 ± √65) / 2
Yes because if 1+0=1 than 0 plus b equals b
I would think -5 plus 5 equals 0.
Assuming the expression you want to solve is: x^2 + 10x + 25 = 81 Subtract both sides by 81: x^2 + 10x - 56=0 First, check for factors of 56 whose difference is 10: 2 and 28, no, 4 and 14, yes. So factor the expression as: (x+14)(x-4)=0 Thus the solutions to the expression are either x=4 or x=-14.
x: x2 - 81 = 0
When the given expression equals 0 then x = -1/6 and x = -6
100x2-81
x2 + 6x + 9 = 81 x2 + 6x = 72 x2 + 6x - 72 = 0 (x+12)(x-6) = 0 x= -12, 6 (two solutions)
That equation can not be factored. It can however be solved for x: x2 - 9x + 4 = 0 x2 - 9x + 81/4 = 81/4 - 4 (x - 9/2)2 = 81/4 - 16/4 (x - 9/2)2 = 65/4 x - 9/2 = ± √65 / 2 x = (9 ± √65) / 2
Yes because if 1+0=1 than 0 plus b equals b
I would think -5 plus 5 equals 0.
Using the discriminant the possible values of k are -9 or 9
A=0 b=0 c=0
No.
x = y = 0 ?