Whether real or complex (Quadratic Equation)

This post deals with the conditions to check whether the roots are real or complex. The roots are real when the graph really cuts the x-axis. When we look at the graph of a quadratic equation we see a cup like structure or cap like structure. Let us check when it is cup like and cap like. We call the cup like structure concave upward and a cap like structure convex upward. The quadratic equation is ax² + bx + c = 0. Let it be equal to y or f(x). Now the rate of change of curve is f'(x) = 2ax + b. f(x) is differentiated to obtain f'(x). We again differentiate f'(x) to obtain f''(x), f''(x) = 2a. Now we find the x-coordinate of the point where the slope is zero i.e. f'(x) = 0.

2ax₁ + b = 0
⇒ x₁ = −b/2aNow the point on the graph is concave upward or convex upward according as the point at the x-coordinate x₁ = −b/2a is maxima or minima. This thing is determined by checking the value of f''(x). If it is positive then the curve has minima and if it is negative then the curve is maxima. We have f''(x) = 2a. It is positive if a is positive and negative when a is negative. Hence we can make following conclusions

1. concave : The graph is concave upward if a is positive.
   
   
   
   The graph given above is of the function f(x) = 2x² + 4x - 3.


2. convex : The graph is convex upward if a is negative.
   
   
   The graph given above is of the function f(x) = −2x² + 4x + 3

Now, we find the y coordinate of the point where it moves upward or downward traveling from x = −∞ at x = -b/2a. It changes its direction at −b/2a to opposite direction of y-axis.
y = ax² + bx + c
at x = x₁ = −b/2a
y₁ = a(−b/2a)² + b(−b/2a) + c
= b²/4a − b²/2a + c
= − b²/4a + c
= − (b² − 4ac)/4a

Now, four conditions arise as given in the following table

a	b2−4ac	Comment
−ve	−ve	The value of y1 is −ve and the curve is concave downward. The curve will be present below y1. Hence it will cut the x-axis at no points. So the roots are imaginary.
−ve	+ve	The value of y1 is +ve and the curve is concave downward. The curve will be present below y1. Hence it will cut the x-axis at one or two points. So the roots are real and may be equal if y1 = 0.
+ve	−ve	The value of y1 is +ve and the curve is concave upward. The curve will be present above y1. Hence it will cut the x-axis at no points. So the roots are imaginary.
+ve	+ve	The value of y1 is −ve and the curve is concave upward. The curve will be present above y1. Hence it will cut the x-axis at one or two points. So the roots are real and may be equal if  y1 = 0.

Quadratic Equation

A quadratic equation is an expression with highest power in it equal to 2 and equated to zero. The term with the power 2 must not have coefficient equal to zero. If the coefficient is equal to zero then the equation is not quadratic anymore. Usually a quadratic equation has three coefficients but when all the coefficients are divided by the coefficients of x² then the coefficients left are two. The three coefficients of a quadratic equation are a,b and c. When they are put in the equation the equation looks like ax² + bx + c = 0.
A quadratic graph looks like this

The equation of the above graph is y = x² + x − 6

When all the terms of the quadratic equation are made negative then the equation looks like below

The equation of the above graph is y = − x² − x + 6

The solution of a quadratic equation by algebra is visible to all. Let us find the solution of the quadratic equation by calculus and graph. As we can observe from the graph that the graph of the curve is symmetrical about the line x = -b/2a. we will prove this when we arrive at the solution of the quadratic equation. When we use calculus to arrive at the solution of a quadratic equation we make use of derivatives. Let y = ax² + bx + c. When the expression is differentiated for the first time then we get ^dy/_dx = 2ax + b. We have got the rate of change of slope as a linear function. Differentiate it again. we get ^d2x/_dx² = 2a. We have found the rate of change of slope of the slope as constant. This implies that the change in slope of the quadratic expression is linear. As the second derivative is positive so the point where ^dy/_dx = 0 is point of minimum value of y. The point is x = ^−b/_2a. Let us shift the y axis to this point. i.e. X = x + b/2a

The equation becomes y = a(X − ^b/_2a)² + b(X − ^b/_2a) + c
y = aX² − bX + ^b²/_4a + bX − ^b²/_2a + c
y = aX² − ^(b²−4ac)/_4a

The roots for the above equation is when y = 0 i.e.
aX² − ^(b²−4ac)/_4a = 0
aX² = ^(b²−4ac)/_4a
X² = ^(b²−4ac)/_4a²
X = ±√(b²−4ac)/2a

As X = x + b/2a
x + b/2a = ±√(b²−4ac)/2a
x = [− b ±√(b²−4ac)]/2a
The two solutions are x₁ and x₂ where x₁ = [− b + √(b²−4ac)]/2a
x₂ = [− b − √(b²−4ac)]/2a
The above method also tells us that the two roots are equidistant from the axis x = −b/2a. This also gives us an idea that the graph is symmetrical about x = −b/2a because the rate of change of slope is constant. And change is slope is linear.