How i can solve quadratic equations which have no real roots
If the quadratic equation has no roots, you cannot find the roots. The discriminate is negative, so if we attempt to use the quadratic equation, we get no roots.
For example, y=x^2 +0x + 3
a=1, b=0, c=3
(-b+/- sqrt(b^2 -4ac))/2a
Substituting in the numbers, we get
x= (-0 +/-sqrt(0^2 - 4(1)3))/2
x= (0 +/- sqrt (-12))/2
We cannot do the square root of -12. Therefore, there are no roots. This is the same as having no x-intercepts. The discriminant is b^2-4ac. In this case it is -12. Thus, there are no real roots.
However, you can still determine the maximum or minimum, the vertex, the axis of symmetry, the y-intercept and the stretch/compression. With this information you can graph the equation.
In this case, the y-intercept is 3, the vertex is at (0,3), the axis of symmetry is x=0, the minimum value of the function is 3.
Let me know if you have any questions.
on Jan 02, 2019