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Anurag yadav
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The Solution of the Quadratic Equation By Differentiation Method
Yes, that can be done. A quadratic equation ##(x\, , \,ax^2+bx+c)## is a parabola. You basically computed where the symmetry axis of a standard parabola lies by determining the x-coordinate of the minimum (##a>0##) or maximum (##a<0##), and then the distance to its two zeros (so they exist). Maybe you are interested to read more about parabolas. https://en.wikipedia.org/wiki/ParabolaAnurag yadav said:The Solution of the Quadratic Equation By Differentiation Method
A quadratic equation is a polynomial equation of the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The solutions to this equation are known as the roots.
Differentiation can be used to find the critical points of the quadratic function f(x) = ax² + bx + c. By finding the derivative f'(x) = 2ax + b and setting it to zero, we can determine the x-coordinate of the vertex, which can help in identifying the nature of the roots.
The vertex of a quadratic equation represents the maximum or minimum point of the parabola. If the vertex lies above the x-axis, the quadratic may have no real roots. If it lies on the x-axis, there is one real root, and if it lies below the x-axis, there are two real roots.
No, the derivative itself does not provide the roots directly. Instead, it indicates the critical points where the function's slope is zero. To find the actual roots, we would still need to solve the original quadratic equation.
Besides differentiation, roots of a quadratic equation can be found using the quadratic formula x = (-b ± √(b² - 4ac)) / (2a), factoring, or by completing the square. Each method has its own advantages depending on the specific equation.