x2 + x - 1 = 0
Using the quadratic formula this gives us:
x = (-1 + or - (-12 - (4 * 1 * -1))1/2) / 2 * 1
x = (-1 + or - (1 - - 4)1/2) / 2
x = (-1 + or - 51/2) / 2
x = 0.618 and -1.618 (Both given to 3 decimal places only)
Note: x1/2 means the square root of x.
x2+7x+3 = 0 Using the quadratic equation formula the solutions are:- x = -6.541381265 or x = -0.4586187349
It has no solutions because the discriminant of the quadratic equation is less than zero.
x = 1 and x = 4
This quadratic equation has no solutions because its discriminant is less than zero.
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+7x+3 = 0 Using the quadratic equation formula the solutions are:- x = -6.541381265 or x = -0.4586187349
There are no real solutions because the discriminant of the quadratic equation is less than zero.
x2+11x+11 = 7x+9 x2+11x-7x+11-9 = 0 x2+4x+2 = 0 The above quadratic equation can be solved by using the quadratic equation formula and it will have two solutions.
x2 + 6x = 16=> x2 + 6x - 16 = 0=> x2 + 8x -2x - 16 = 0=> (x+8)(x-2) = 0=> x = -8 or x = 2So, the solutions of the quadratic equation x2 + 6x = 16 are -8 and 2.
It has no solutions because the discriminant of the quadratic equation is less than zero.
x = 1 and x = 4
This quadratic equation has no solutions because its discriminant is less than zero.
x2 + 49 = 0
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+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
None because without an equal it is not an equation. But if it was in the form of x2+7x+12 = 0 then it would have 2 solutions which are x = -3 and x = -4
There are none. For this equation, there is nonreal answer, as the graph of the quadratic does not pass below the x-axis