Integral of sqrt(2x) = integral of (2x)1/2 = √2/(3/2)*(x)3/2 + c = (2√2)/3*(x)3/2 + c where c is the constant of integration. Check: ( (2√2)/3*(2x)3/2 + c )' = (2√2)/3*(3/2)(x)(3/2)-1 + 0 = √2*(x)1/2 = √2x
6x + 9 = -3 6x/3 + 9/3 = -3/3 2x + 3 = -1 2x + 3 - 3 = -1 - 3 2x = -4 2x/2 = -4/2 x = -2
4x2-4x-3 = 0 (2x-3)(2x+1) = 0 x = 3/2 or x = -1/2
3x + 5x = 2x + 28x = 2x + 26x = 2x = (2/6) = 1/3
if you mean 2x-1*3 then the answer is: 2x-3
f(x) = √(2x -3) f(x) = (2x - 3)^(1/2) f'(x) = (1/2)[(2x - 3)^(1/2 - 1)](2) f'(x) = (2x - 3)^(-1/2) f'(x) = 1/[(2x - 3)^(1/2)] f'(x) = 1/√(2x -3)
3+4x+2-2x=2x+5
x = 1 ------------- 3 = 2x + 1 -1 -1 2=2x /2 /2 1=x
3
2x+1 because -2 +3 = 1
4x2 - 2x - 3 = 0 (2x + 1)(2x - 3) = 0 x = -(1/2) or (3/2)
(2X/3) = 1multiply through by 33[(2X/3) = 1]2X = 3divide both sides integers by 2(2/2)X = 3/2X = 3/2========check in original equation2(3/2)/3 = 1(6/2)/3 = 1 ( or just cancel 2 in numerator and denominator )3/3 = 11 = 1=====checks
If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2
2x-1=-32x=-3+1x=-2/2x=-1
If: 2x-1 = 2Then: 2x = 2+1And: 2x = 3So: x = 3/2 = 1.5 or 1 and a 1/2
2x +1 = 5x-2 2x +3= 5x 3 = 3x x=1
(2x + 3)(x + 1 ) = 2x^2 +5x +3