You must add 1
x2 - 2x - 5 = 0 x2 - 2x + 1 = 6 (x - 1)2 = 6 x - 1 = ± √6 x = 1 ± √6
x=1
A graph that has 1 parabolla that has a minimum and 1 positive line.
x=5.54x+1= 2x+12-1 -14x=2x+11-2x -2x2x=11/2 /2x=5.5
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
The vertex is at (-1,0).
x2+10x+1 = -12+2x x2+10x-2x+1+12 = 0 x2+8x+13 = 0 Solving by using the quadratic equation formula: x = - 4 - or + the square root of 3
4X + 2x = 1. Where x = 0.166666666666666666666666666666666666666 recurring.
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
x2 - 10x - 6 = 2x + 1 Subtracting 2x + 1 from both sides gives: x2 - 12x - 7 = 0
x2+10x+1 = -12+2x x2+8x+13 = 0 The solution: x = -4 + the square root of 3 or x = -4 - the square root of 3
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
x2 + 2x - 38 = 0 ∴ x2 + 2x + 1 = 39 ∴ (x + 1)2 = 39 ∴ x + 1 = ±√39 ∴ x = -1 ±√39
x2+2x+1 = 0 (x+1)(x+1) = 0 x = -1 and also x = -1
If: x2+2x-3 = 0 Then: x = 1 or x = -3