Quadratic Formula: Derivation and Worked Examples

The quadratic formula is a fundamental concept in algebra and calculus. It allows us to solve any quadratic equation of the form ax^2 + bx + c = 0, where a, b, and c are constants, and x is an unknown variable. In this article, we will derive the quadratic formula from scratch, showing every step of the process with clear explanations and worked examples. We will also discuss common mistakes to avoid and how to check solutions by substitution.

The Quadratic Equation

A quadratic equation is an equation of the form:

\[ax^2 + bx + c = 0\]

where a, b, and c are constants, and x is an unknown variable. In this article, we will assume that a ≠ 0. The number a is called the leading coefficient of the quadratic equation, while b and c are called the linear coefficient and constant term, respectively.

To solve a quadratic equation, our goal is to find values of x that make the left-hand side equal to zero:

\[ax^2 + bx + c = 0\]

Derivation of the Quadratic Formula

We will now derive the quadratic formula using elementary algebraic manipulations. Our final goal is to find an expression for x in terms of a, b, and c. To begin, we can rewrite the equation as follows:

[(ax^2 + bx + c) - c = 0] [\text{or}] [ax^2 + bx = -c]

Now, let's factor out an x from the left-hand side:

\[x(a x + b) = -c\]

Next, we can divide both sides by (a x + b), assuming that a x + b ≠ 0. This step is valid because dividing a polynomial by another nonzero polynomial results in the quotient being:

\[\frac{ax^2 + bx}{a x + b} = \frac{ax^2}{a x + b} + \frac{bx}{a x + b} = x\]

So, our equation now becomes:

\[x = -\frac{c}{a x + b}\]

Finally, we can solve for x by isolating the variable on one side of the equation and rearranging terms. To do this, multiply both sides by (a x + b), which eliminates the fraction:

[ax^2 + bx = -c] [\text{or}] [ax^2 + bx + c = 0]

Now, we have returned to our original quadratic equation. We can now read off the solution for x by substituting in the expression we found earlier:

\[x = -\frac{c}{a x + b}\]

Therefore, the general solution for a quadratic equation is given by the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula applies to any quadratic equation of the form ax^2 + bx + c = 0.

Examples and Applications

Example 1: Solving a Quadratic Equation Using the Quadratic Formula

Consider the quadratic equation 3x^2 - 4x + 7 = 0. We will solve this equation using the quadratic formula. First, we identify the coefficients as follows:

[a = 3,] [b = -4,] and [c = 7.]

Next, we plug these values into the quadratic formula:

\[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(7)}}{2(3)}\]

Simplifying, we get:

[x = \frac{4 \pm \sqrt{16 - 504}}{6}] [x = \frac{4 \pm \sqrt{-488}}{6}]

Since the discriminant (the expression under the square root) is negative, we know that this equation has no real solutions. Therefore, the solution set for 3x^2 - 4x + 7 = 0 consists only of complex numbers:

\[x = \frac{4 \pm i\sqrt{488}}{6}\]

Example 2: Solving a Quadratic Equation by Factoring

Consider the quadratic equation 2x^2 + x - 3 = 0. We will solve this equation by factoring. First, we notice that the factors of -3 are ±1 and ±3, so our equation can be factored as:

\[2x^2 + x - 3 = (x - 1)(2x - 3) = 0\]

Now, setting each factor equal to zero gives us the solutions:

[(x - 1) = 0 \Rightarrow x = 1] [(2x - 3) = 0 \Rightarrow x = \frac{3}{2}]

Thus, we have two real solutions for this quadratic equation: x = 1 and x = 3/2.

Common Mistakes and Substitution Checking

When solving a quadratic equation, it is essential to check your solutions by substituting them back into the original equation. This step ensures that you have found all valid solutions and helps catch any errors in your calculations.

For instance, if we had mistakenly computed the solutions for Example 1 as:

[x = \frac{4 \pm \sqrt{488}}{6}] [x = \frac{4 \mp i\sqrt{488}}{6}]

We would have made an error in our computation of the square root, resulting in complex solutions. To catch this mistake, we could substitute these solutions back into the original equation:

\[3\left(\frac{4 \pm i\sqrt{488}}{6}\right)^2 - 4\left(\frac{4 \pm i\sqrt{488}}{6}\right) + 7 = 0\]

Upon simplification, we would find that the left-hand side does not equal zero. This indicates an error in our calculations and suggests that we should revisit the computation of the square root.

Conclusion

In this article, we have derived the quadratic formula and explored its applications through two examples: one using the formula itself and another by factoring. We also discussed the importance of checking solutions by substitution to ensure their validity. The quadratic formula is a powerful tool for solving any quadratic equation, providing valuable insights into the nature of parabolas and their graphical representations in physics, chemistry, and other scientific disciplines.