x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
Assuming the 2 is meant to be a square this is the form: x2 - 3x2 = -2x2 + 8x + 5
3x2 + 2x = 16 ∴ 3x2 + 2x - 16 = 0 ∴ 3x2 - 6x + 8x - 16 = 0 ∴ 3x(x - 2) + 8(x - 2) = 0 ∴ (3x + 8)(x - 2) = 0 ∴ x ∈ {-8/3, 2}
The answe is 27x^4 + 6x^3 - 61x^2 - 68x - 42
Factors are (3x + 2)(x + 2) so x is either -2/3 or -2 If term in x is negative then factors are (3x - 2)(x - 2) and solutions are 2/3 and 2.
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
The sum 6 (3x^2) + 8x -3 plus 3 (3x^2) - 7x + 2 = 27 x^2 + x -1
(3x + 2)(x + 2) so x = -2/3 or -2
2x^2 + 8x + 3 = 0
Assuming the 2 is meant to be a square this is the form: x2 - 3x2 = -2x2 + 8x + 5
3x2 + 2x = 16 ∴ 3x2 + 2x - 16 = 0 ∴ 3x2 - 6x + 8x - 16 = 0 ∴ 3x(x - 2) + 8(x - 2) = 0 ∴ (3x + 8)(x - 2) = 0 ∴ x ∈ {-8/3, 2}
3x2 + 8x - 8 doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (-4 plus or minus 2 times the square root of 10) divided by three.x = 0.7748517734455863x = -3.441518440112253
3x-3x+4x^2+8x=4x^2+8x
The answe is 27x^4 + 6x^3 - 61x^2 - 68x - 42
3x^2 + 8x + 4 = (x+2) (3x+2) x = -2 x = -2/3 So there are no complex roots, they are real. You can test this by b^2 - 4ac if greater than 0, it is real if equal, there will be 2 identical roots. if less than 0 you get imaginary roots.
Factors are (3x + 2)(x + 2) so x is either -2/3 or -2 If term in x is negative then factors are (3x - 2)(x - 2) and solutions are 2/3 and 2.
3x2+26x+16 When factorised: (3x+2)(x+8)