Solving Quadratic Equations with Quadratic Formula

Solving Quadratic Equations

Quadratic equations are equations of the form (ax^2 + bx + c = 0), where (a), (b), and (c) are constants, and (x) represents the variable. Solving quadratic equations is a fundamental part of algebra, and it's a skill you'll often encounter in mathematics. In this article, we'll dive into the quadratic formula, a method for finding the solutions (roots) of quadratic equations.

The Quadratic Formula

The quadratic formula provides a general method for solving quadratic equations. It's based on the fact that every quadratic equation has two roots, which are the solutions to the equation. The formula is:

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

This formula works for any quadratic equation (ax^2 + bx + c = 0), provided that (a \neq 0).

Let's break down the formula:

• The variable (x) is the root we're trying to find.
• The (-b) term represents half of the coefficient of (x) in our original equation.
• The (\sqrt{b^2 - 4ac}) term represents the distance between the two roots, which is the same as the magnitude of the discriminant.
• The (2a) term is a scaling factor that ensures the formula gives us valid roots.

The Discriminant

The discriminant, denoted as (b^2 - 4ac), is a key component of the quadratic formula. The discriminant determines whether our quadratic equation has real roots, complex roots, or no roots.

• If the discriminant is positive, the equation has two distinct real roots.
• If the discriminant is zero, the equation has only one real root.
• If the discriminant is negative, the equation has no real roots; instead, it has complex roots, which occur in conjugate pairs.

You can use the discriminant to classify the roots of a quadratic equation, and to help you decide which form of the quadratic formula to use.

The Two Forms of the Quadratic Formula

When solving quadratics, you may encounter two different forms of the quadratic formula:

1. The standard form: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a})
2. The factored form: (x = \frac{-b \pm \sqrt{b^2 - k}}{2}) where (k = ac).

You can use these two forms interchangeably, provided that you convert between the two appropriately.

Practice Problems

Let's solve a few examples to practice using the quadratic formula:

1. Solve the equation (x^2 + 3x + 2 = 0).

Using the standard form:

[x = \frac{-3 \pm \sqrt{3^2 - 4(1)(2)}}{2(1)} = \frac{-3 \pm \sqrt{9 - 8}}{2} = \frac{-3 \pm \sqrt{1}}{2} = \frac{-3 \pm 1}{2}]

So the solutions are (x = \frac{-3 + 1}{2} = \frac{-2}{2} = -1) or (x = \frac{-3 - 1}{2} = \frac{-4}{2} = -2).

2. Solve the equation (2x^2 - 5x + 3 = 0).

Using the standard form:

[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(3)}}{2(2)} = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm \sqrt{1}}{4} = \frac{5 \pm 1}{4}]

So the solutions are (x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}) or (x = \frac{5 - 1}{4} = \frac{4}{4} = 1).

3. Solve the equation (x^2 - 4x + 1 = 0).

Using the standard form:

[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2}]

The discriminant is 12, which is positive, so the equation has two distinct real roots. We can simplify the expression further by taking the square root of 12:

[\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}]

So the solutions are (x = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}) or (x = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}).

Summary

Solving quadratic equations using the quadratic formula is a fundamental skill in algebra. The formula provides a general method for finding the roots of quadratic equations. Understanding the discriminant is essential for classifying the roots of a quadratic equation and determining which form of the quadratic formula to use. Practice problems will help you gain confidence and mastery in solving quadratic equations.