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.
A quadratic graph looks like this
When all the terms of the quadratic equation are made negative then the equation looks like below
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
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.
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