Solving Quadratic Equations by Quadratic Formula

Solving Quadratics by the Quadratic Formula

In this lesson, we will learn how to solve quadratic equations using the quadratic formula. Quadratic equations are equations of the form:

ax² + bx + c = 0

The quadratic formula provides a way to find the solutions to any quadratic equation:

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

Lesson Outline:

• What is the Quadratic Formula?
• Steps to Solve Quadratics Using the Formula
• 6 Example Problems
• Practice Exercises with Solutions
• Homework Section
• Revision Section

Steps to Solve Quadratics Using the Quadratic Formula

1. Write the quadratic equation in standard form: $ax^2 + bx + c = 0$
2. Identify the values of a, b, and c
3. Substitute these values into the quadratic formula
4. Simplify under the square root (the discriminant)
5. Calculate the solutions

Examples

Example 1:

Solve $2x^2 + 3x - 5 = 0$ using the quadratic formula.

Solution:

Here, $a = 2$, $b = 3$, and $c = -5$.

Using the quadratic formula:

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

$x = \frac{-3 \pm \sqrt{9 + 40}}{4}$

$x = \frac{-3 \pm \sqrt{49}}{4}$

$x = \frac{-3 \pm 7}{4}$

Therefore, the solutions are $x = 1$ and $x = -2.5$.

Example 2:

Solve $x^2 - 4x - 8 = 0$ using the quadratic formula.

Solution:

Here, $a = 1$, $b = -4$, and $c = -8$.

Using the quadratic formula:

$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-8)}}{2(1)}$

$x = \frac{4 \pm \sqrt{16 + 32}}{2}$

$x = \frac{4 \pm \sqrt{48}}{2}$

Simplify the square root of 48: $x = \frac{4 \pm 4\sqrt{3}}{2}$

The solutions are $x = 2 + 2\sqrt{3}$ and $x = 2 - 2\sqrt{3}$.

Exercises

Exercise 1:

Solve $3x^2 + 6x - 9 = 0$ using the quadratic formula.

Exercise 2:

Solve $x^2 + 2x - 3 = 0$ using the quadratic formula.

Solutions to Exercises

Solution 1:

Here, $a = 3$, $b = 6$, and $c = -9$.

Using the quadratic formula:

$x = \frac{-6 \pm \sqrt{6^2 - 4(3)(-9)}}{2(3)}$

$x = \frac{-6 \pm \sqrt{36 + 108}}{6}$

$x = \frac{-6 \pm \sqrt{144}}{6}$

$x = \frac{-6 \pm 12}{6}$

The solutions are $x = 1$ and $x = -3$.

Solution 2:

Here, $a = 1$, $b = 2$, and $c = -3$.

Using the quadratic formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)}$

$x = \frac{-2 \pm \sqrt{4 + 12}}{2}$

$x = \frac{-2 \pm \sqrt{16}}{2}$

$x = \frac{-2 \pm 4}{2}$

The solutions are $x = 1$ and $x = -3$.

Homework

Solve the following quadratic equations using the quadratic formula:

• $2x^2 + 5x + 3 = 0$
• $x^2 - 6x + 8 = 0$
• $4x^2 + 4x + 1 = 0$

Revision

Remember to always check if the quadratic equation is in standard form before applying the quadratic formula. Pay attention to the signs of the coefficients and simplify the discriminant carefully.

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