negative b plus or minus the square root of b minus 4 times ac all over 2 a
x equals negative b plus or minus the square root of b squared minus 4bc over 2a
In a quadratic equation, the vertex (which will be the maximum value of a negative quadratic and the minimum value of a positive quadratic) is in the exact center of any two x values whose corresponding y values are equal. So, you'd start by solving for x, given any y value in the function's range. Then, you'd solve for y where x equals the middle value of the two x's given in the previous. For example:y = x24 = x2x = 2, -2y = (0)2y = 0Which is, indeed, the vertex of y = x2
Where the equation is ax2 + bx + c the roots are given by the solutions to : (-b +/- sqrt(b2 - 4ac))/2a So you would first input the values of a, b , and c, then calculate [d = b*b - 4*a*c], and then determine whether it is positive or negative. If it is positive [d>0, or d=0], then you can continue: so the two roots of x are: (-b + sqrt(d))/2/a, and (-b - sqrt(d))/2/a. If d < 0, then you need to calculate sqrt(-d) and then display -b/2/a as the real part, and +/- sqrt(-d)/2/a as the imaginary parts.
The quadric equation is: negative b plus or minus the square root of b squared minus 4ac all over(divided by) 2a
In basic mathematics, a quadratic equation with a negative discriminant has no solutions. However, at a more advanced level you will learn that it has two solutions which form a complex conjugate pair.
By calculating the discriminant of the equation and if it's negative the equation will have no solutions
There are two complex solutions.
No. By definition, a quadratic equation can have at most two solutions. For a quadratic of the form ax^2 + bx + c, when the discriminant of a quadratic, b^2 - 4a*c is positive you have two distinct real solutions. As the discriminant becomes smaller, the two solutions move closer together. When the discriminant becomes zero, the two solutions coincide which may also be considered a quadratic with only one solution. When the discriminant is negative, there are no real solutions but there will be two complex solutions - that is those involving i = sqrt(-1).
If the discriminant is negaitve, there are no "real" solutions. The solutions are "imaginary".
As stated in the attached link, there are three possible discriminant conditions: Positive, Zero, or Negative. If the discriminant is negative, there are no real solutions but there are two imaginary solutions. So, yes there are solutions if the discriminant is negative. The solutions are imaginary, which is perfectly acceptable as solutions.
The discriminant tells you how many solutions there are to an equation The discriminant is b2-4ac For example, two solutions for a equation would mean the discriminant is positive. If it had 1 solution would mean the discriminant is zero If it had no solutions would mean that the discriminant is negative
It has a complete lack of any x-intercepts.
It has two complex solutions.
Assuming the coefficients are real, the discriminant is non-negative. The reason for this is that in this case, if the solutions are complex, they must needs be conjugats of one another, i.e., two different solutions.