akira888 said:
Um, the square of 5 + (3x+2)^0.5 is 25 + 10 * sqrt(3x+2) + (3x+2), not 25 + (3x+2)
(a+b)^2 = a^2 + 2*a*b + b^2
(a+b)^2 != a^2 + b^2
oops, my bad... (hangs head in shame)
=Xenus]sqrt(3 - 3x) * sqrt(3 - 3x) = 3 + ( sqrt(3) * sqrt(3x) ) + ( sqrt(3) * sqrt(3x) ) + 3x
should be
sqrt(3 - 3x) * sqrt(3 - 3x) = 3 + ( sqrt(3) * sqrt(-3x) ) + ( sqrt(3) * sqrt(-3x) ) + 3x
sqrt(3-3x) != sqrt(3) -sqrt(3x), so both are wrong.
Been years since I've had to do any sort of quadratic algebra, the rust is showing. Anyway here's my latest stab at working it out:
sqrt(3-3x) = 5+ sqrt(3x+2)
[squared both sides of the eqn]
(3-3x) = 25 + 2*5*sqrt(3x+2) + (3x+2)
[re-arrange]
10*sqrt(3x+2) = -(6x+24)
[square again]
100*(3x+2) = 36*(x²+8x+16)
[re-arrange]
x² + 8x - (100/12)x + 16 - (100/18) = 0
x² - (1/3)x + (94/9) = 0
eqn is of the form ax²+bx+c, using the quadratic formula
x = -b ± (sqrt(b² - 4*a*c))÷(2*a)
but because (b² - 4*a*c) here gives us a negative result the roots are complex.