Quadratic Formula Explained

Any quadratic equation of the form ax² + bx + c = 0 can be solved using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

This formula allows us to find the two possible values of x that make the equation true. A quadratic equation can have up to two distinct solutions, depending on the value of the expression under the square root.

Example: Solving a quadratic equation step by step

Let’s solve the equation:

$$ 2x^2 + 5x + 3 = 0 $$

Here the coefficients are a = 2, b = 5, and c = 3. Substitute these values into the formula:

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

$$ x = \frac{-5 \pm \sqrt{5^2 - 4(2)(3)}}{2(2)} $$

$$ x = \frac{-5 \pm \sqrt{25 - 24}}{4} $$

$$ x = \frac{-5 \pm \sqrt{1}}{4} $$

$$ x = \frac{-5 \pm 1}{4} $$

That gives us two solutions:

$$ x = \begin{cases} \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2} \\ \\ \frac{-5 + 1}{4} = \frac{-4}{4} = -1 \end{cases} $$

So, the equation has two roots: x₁ = -3/2 and x₂ = -1.

Checking the solutions

Verification. Let’s plug the first value x₁ = -3/2 into the equation: $$ 2x^2 + 5x + 3 = 0 $$ $$ 2(-\frac{3}{2})^2 + 5(-\frac{3}{2}) + 3 = 0 $$ $$ 2(\frac{9}{4}) - \frac{15}{2} + 3 = 0 $$ $$ \frac{9}{2} - \frac{15}{2} + 3 = 0 $$ $$ \frac{-6}{2} + 3 = 0 $$ $$ -3 + 3 = 0 $$ $$ 0 = 0 $$ Now check the second value x₂ = -1: $$ 2x^2 + 5x + 3 = 0 $$ $$ 2(-1)^2 + 5(-1) + 3 = 0 $$ $$ 2 - 5 + 3 = 0 $$ $$ 0 = 0 $$ Both results confirm the correctness of our solutions.

How the quadratic formula is derived

To understand where the quadratic formula comes from, let’s start from the general form:

$$ ax^2 + bx + c = 0 $$

Subtract c from both sides to isolate the terms with x:

$$ ax^2 + bx = -c $$

Divide through by a to simplify:

$$ x^2 + \frac{b}{a}x = -\frac{c}{a} $$

Now, to complete the square, take half of the coefficient of x, square it, and add it to both sides:

$$ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 $$

The left side is now a perfect square:

$$ \left(x + \frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2} $$

Rewrite the right-hand side with a common denominator:

$$ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} $$

Take the square root of both sides:

$$ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} $$

Finally, isolate x:

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

The expression under the square root, b² - 4ac, is called the discriminant. It determines the nature of the solutions:

• If b² - 4ac > 0, there are two distinct real roots.
• If b² - 4ac = 0, there is one repeated real root.
• If b² - 4ac < 0, the roots are complex.

 

This derivation shows not only how to find the solutions but also why the quadratic formula works for every second-degree equation.