x3 + x2 - x = 0
∴ x2 + x - 1 = 0
∴ x2 + x + 0.25 = 1.25
∴ (x + 0.5)2 = 1.25
∴ x + 0.5 = ± 1.251/2
∴ x = -0.5 ± 1.251/2
Realising that 125 is a perfect cube, this is actually pretty easy: x3 + 125 = 0 x3 = -125 x = -5
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
IMPOSSIBLE
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
No.
Realising that 125 is a perfect cube, this is actually pretty easy: x3 + 125 = 0 x3 = -125 x = -5
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
IMPOSSIBLE
cube root of 216 = 6 > X= 6
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
No.
y2=x3+3x2
No, it is not.
It is x = -5
(xn+2-1)/(x2-1)
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0