(x-6)(x+4) = 0 x = 6 or x = -4
If: x2-4 = 0 Then: x = 2
I will assume you mean,X2 - 16 = 0X2 = 16take square root each sideX = (+/-) 4=========(-4, 0) and (4, 0)----------------------
-4 and 4
x2 - 6x + 8 = 0 x2 - 4x - 2x + 8 = 0 x*(x - 4) - 2*(x - 4) = 0 (x - 2)*(x - 4) = 0 so (x - 2) = 0 or (x - 4) = 0 ie x = 2 or x = 4
x2 - 4x = 0 x(x - 4) = 0 so either x = 0 or x - 4 = 0 that is, x = 0 or x = 4.
(x-6)(x+4) = 0 x = 6 or x = -4
If: x2-4 = 0 Then: x = 2
I'll show work. X2-8x-4=0. X2-8x=4. X-8x=2. -7x=2. X=-(2/7).
I will assume you mean,X2 - 16 = 0X2 = 16take square root each sideX = (+/-) 4=========(-4, 0) and (4, 0)----------------------
-4 and 4
x2 - 6x + 8 = 0 x2 - 4x - 2x + 8 = 0 x*(x - 4) - 2*(x - 4) = 0 (x - 2)*(x - 4) = 0 so (x - 2) = 0 or (x - 4) = 0 ie x = 2 or x = 4
have to put it was 2x and not x2--back to the problem--2x-x-4=0x=4* * * * *True, but the question, in all likelihood WAS x2 - x - 4 = 0and the solutions to that equation are -1.561553 and 2.561553
If you mean to say (x^2)-2x+4=0 then there is no solution for x because the equation never makes it to 0 If you mean to say 2x-2x+4=0 then there is no solution for x because it is a horizontal line at Y=4
X2 - 3X - 4 = 0factored(X + 1)(X - 4)----------------X = - 1X = 4-------------------(- 1, 0) and (4, 0)================solution set of X interception points
x2 - 4x + 4 = 0 (x - 2)(x - 2) = 0 x = 2
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4