There are no real solutions to this equation because you cannot take the square root of a negative number. However,x2 + 4 = 0x2 = -4sqrt(x2) = sqrt(-4)x = 2i, -2ihere are the imaginary solutions.
x2 + 3x + 3 = 0 doesn't factor neatly. Applying the quadratic formula we find two imaginary solutions: (-3 plus or minus the square root of -3) divided by 2x = -1.5 + 0.8660254037844386ix = -1.5 - 0.8660254037844386iwhere i is the square root of negative 1
Using the quadratic equation formula: x = -3 - the square root of 3 or x = -3 + the square root of 3
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There are no real solutions to this equation because you cannot take the square root of a negative number. However,x2 + 4 = 0x2 = -4sqrt(x2) = sqrt(-4)x = 2i, -2ihere are the imaginary solutions.
Yes. For example, the equation x2 = 2, which in standard form is x2 - 2 = 0, has the two solutions x = square root of 2, and x = minus square root of 2.
x2 + 3x + 3 = 0 doesn't factor neatly. Applying the quadratic formula we find two imaginary solutions: (-3 plus or minus the square root of -3) divided by 2x = -1.5 + 0.8660254037844386ix = -1.5 - 0.8660254037844386iwhere i is the square root of negative 1
x2 + 3x - 18 = 0 (x - 3)(x + 6) = 0 x = 3 or -6
Using the quadratic equation formula: x = -3 - the square root of 3 or x = -3 + the square root of 3
X2+11x+11 = 7x+9 X2+11x-7x+11-9 = 0 x2+4x+2 = 0 Solve as a quadratic equation by using the quadratic equation formula or by completing the square: x = -2 + or - the square root of 2
x^2 + 3x +7 = 0 doesn't factor neatly. Applying the quadratic equation, we find two imaginary solutions: (-3 plus or minus the square root of -19) divided by 2x = -1.5 + 2.179449471770337ix = -1.5 - 2.179449471770337iwhere i is the square root of negative 1
It is the equation inside the square root of the Quadratic FormulaIf > 0 there is a solutionIf < 0 there is no solutionBecause you can not calculate the Square Root of a Negative Number
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x2 - 2x - 15 = 0
x2-10 = 0 x2 = 10 x = the square root of 10