Lesson 5: Solving Quadratic Equations Using the Quadratic Formula

Property

Before we use the quadratic formula, we must write the equation in standard form, $ax^2 + bx + c = 0$, so that we can identify the coefficients $a$, $b$, and $c$. If the coefficients are fractions, it helps to clear the fractions by multiplying each term by the LCD.

Examples

• To solve $5x^2 = 8x - 3$, first rearrange it to standard form: $5x^2 - 8x + 3 = 0$. Now you can correctly identify $a=5, b=-8,$ and $c=3$.

• Given the equation $x^2 - \frac{x}{3} = \frac{4}{3}$, first multiply by the LCD, 3, to get $3x^2 - x = 4$. Then, rewrite in standard form as $3x^2 - x - 4 = 0$, so $a=3, b=-1, c=-4$.

Property

The solutions to a quadratic equation of the form $ax^2 + bx + c = 0$, where $a \neq 0$ are given by the formula:

To solve a quadratic equation using the Quadratic Formula:
Step 1. Write the quadratic equation in standard form, $ax^2 + bx + c = 0$. Identify the values of $a$, $b$, and $c$.
Step 2. Write the Quadratic Formula. Then substitute in the values of $a$, $b$, and $c$.
Step 3. Simplify.
Step 4. Check the solutions.

Examples

• To solve $2x^2 + 5x - 3 = 0$, we identify $a=2, b=5, c=-3$. Substituting into the formula gives $x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}$. The solutions are $x = \frac{1}{2}$ and $x = -3$.
• To solve $3x^2 + 10x + 5 = 0$, we have $a=3, b=10, c=5$. The formula gives $x = \frac{-10 \pm \sqrt{10^2 - 4(3)(5)}}{2(3)} = \frac{-10 \pm \sqrt{100 - 60}}{6} = \frac{-10 \pm \sqrt{40}}{6} = \frac{-10 \pm 2\sqrt{10}}{6} = \frac{-5 \pm \sqrt{10}}{3}$.
• To solve $x^2 + 2x + 10 = 0$, we have $a=1, b=2, c=10$. The formula gives $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(10)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 40}}{2} = \frac{-2 \pm \sqrt{-36}}{2} = \frac{-2 \pm 6i}{2} = -1 \pm 3i$.

Explanation

The Quadratic Formula is a powerful tool derived from completing the square on the general quadratic equation. It provides a direct solution for any quadratic equation, saving you from repeating the steps of completing the square every time.

Property

In the Quadratic Formula, $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, the quantity $b^2 - 4ac$ is called the discriminant.
For a quadratic equation of the form $ax^2 + bx + c = 0$, $a \neq 0$: - If $b^2 - 4ac > 0$, the equation has 2 real solutions. - If $b^2 - 4ac = 0$, the equation has 1 real solution. - If $b^2 - 4ac < 0$, the equation has no real solutions.

Examples