Proof of the Quadratic Formula Using Differentiation
Given the quadratic equation: \[ ax^2 + bx + c = 0 \]
From the given equation, we have: \[ ax^2 + bx = -c \]
Let \( y = ax^2 + bx + c \).
Differentiating \( y \) with respect to \( x \): \[ \frac{dy}{dx} = 2ax + b \]
Now, let's consider the square of \( \frac{dy}{dx} \): \[ \left(\frac{dy}{dx}\right)^2 = (2ax + b)^2 \] \[ \left(\frac{dy}{dx}\right)^2 = 4a^2x^2 + 4abx + b^2 \]
We can rewrite the expression by factoring out \( 4a \) from the first two terms: \[ 4a(ax^2 + bx) + b^2 \]
From the initial given equation, we know that \( ax^2 + bx = -c \). Substitute this into the expression: \[ 4a(-c) + b^2 = -4ac + b^2 = b^2 - 4ac \]
So, we have: \[ \left(\frac{dy}{dx}\right)^2 = b^2 - 4ac \]
Taking the square root of both sides: \[ \frac{dy}{dx} = \pm\sqrt{b^2 - 4ac} \]
We also know that \( \frac{dy}{dx} = 2ax + b \). Therefore, we can equate the two expressions for \( \frac{dy}{dx} \): \[ 2ax + b = \pm\sqrt{b^2 - 4ac} \]
Now, we want to solve for \( x \). First, subtract \( b \) from both sides: \[ 2ax = -b \pm\sqrt{b^2 - 4ac} \]
Finally, divide by \( 2a \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
