x2+2x+2=0
Using the quadratic formula
x= (-2+- sq.root (22-4.1.2))/2
x= (-2+- sq.root(4-8))/2
x= (-2+- sq.root -4)/2
x= (-2+- 2i)/2
x= -2+2i and -2-2i which then simplifies to -1+i and -1-i.
Hope this wasn't too confusing.
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
If: x2+2x-3 = 0 Then: x = 1 or x = -3
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
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
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
If: x2+2x-3 = 0 Then: x = 1 or x = -3
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
y = x2 + 2x + 1zeros are:0 = x2 + 2x + 10 = (x + 1)(x + 1)0 = (x + 1)2x = -1So that the graph of the function y = x + 2x + 1 touches the x-axis at x = -1.
x2 + 2x - 13 = 2 x2 + 2x - 15 = 0 (x + 5)(x - 3) = 0 x + 5 = 0 and x - 3 = 0 so x = -5 and x = 3
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
2x^2 + 8x + 3 = 0
No real roots
x2 + 2x - 15 = 0(x + 5) (x - 3) = 0x = -5x = +3
x2 + 2x - 38 = 0 ∴ x2 + 2x + 1 = 39 ∴ (x + 1)2 = 39 ∴ x + 1 = ±√39 ∴ x = -1 ±√39
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)