Factors of 5 are only 1 and 5, which cannot lead to 2x, so using the quadratic formula (-2 +/- sqrt(4 - 20))/2 = (-2 +/- 4i))/2 = -1 +/- 2i which takes us into the realm of imaginary numbers...
x2+12x+36=0 2x+12x+36=0 14x+36=0 14x=-36 x=-36/14 x=-2 4/7
x²+2x-15=0 x1=-2/2 - Square root of ((2/2)²+15) x1=-1-4 x1=-5 x2=-2/2 + Square root of ((2/2)²+15) x2=-1+4 x2=3
x2+2x-4 = 0 x = (-2 plus/minus sqrt (4-(4x1x-4)) / 2 x = (-2 plus/minus sqrt 20)/2 x = (-2 - sqrt 20)/2 or (-2 + sqrt 20)/2 x = -3.236 or 1.236
x3 + x2 - 8x - 6 = 0 x3 + 3x2 - 2x2 - 6x - 2x - 6 = 0 x2(x + 3) - 2x(x + 3) - 2(x + 3) = 0 (x + 3)(x2 - 2x - 2) = 0 The second bracket cannot be factorised, so applying the quadratic formula to it gives the answers as So x = -3 or x = 1 +/-sqrt(3) that is, x = -3, x = -0.732051 or x = 2.732051
x2 + 36 = 20x so x2 - 20x + 36 = 0 so x2 - 2x - 18x + 36 = 0 or x(x - 2) - 18(x - 2) = 0 that is (x - 2)(x - 18) = 0 so x - 2 = 0 or x - 18 = 0 ie x = 2 or x = 18
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)