Lesson 2.5: Quadratic Equations

New Concept

A quadratic equation involves a variable squared, like $x^2$. This lesson equips you with four essential methods to solve them: factoring, using square roots, completing the square, and applying the powerful quadratic formula, $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

What’s next

You've got the big picture. Next, you'll dive into practice cards and short videos that break down each of the four solving methods, starting with factoring.

Property

Factoring involves writing a quadratic equation as a product of linear terms. This method relies on the zero-product property, which states that if $a \cdot b = 0$, then $a = 0$ or $b = 0$. The equation must be in standard form $ax^2 + bx + c = 0$.

When a = 1: Find two numbers whose product is $c$ and whose sum is $b$. Use them to create factors $(x+k_1)(x+k_2)=0$.

When a ≠ 1: Use grouping. Find two numbers whose product is $ac$ and sum is $b$. Rewrite the $bx$ term using these numbers, then factor the resulting four-term polynomial.

Property

With the $x^2$ term isolated, the square root property states that: if $x^2 = k$, then $x = \pm \sqrt{k}$ where $k$ is a nonzero real number.

To solve using this method:

1. Isolate the $x^2$ term on one side of the equal sign.
2. Take the square root of both sides of the equation, putting a $\pm$ sign before the expression on the side opposite the squared term.
3. Simplify the numbers on the side with the $\pm$ sign.

Examples

• To solve $x^2 = 121$, we take the square root of both sides: $x = \pm\sqrt{121}$, so the solutions are $x=11$ and $x=-11$.

Property

Completing the square is a method for solving a quadratic equation by creating a perfect square trinomial on one side. Given an equation in the form $x^2 + bx = c$, we can transform it:

1. Multiply the $b$ term by $\frac{1}{2}$ and square it: $(\frac{b}{2})^2$.
2. Add this value to both sides of the equation: $x^2 + bx + (\frac{b}{2})^2 = c + (\frac{b}{2})^2$.
3. The left side now factors to $(x + \frac{b}{2})^2$, allowing you to solve using the square root property.

Examples

• Solve $x^2 + 10x = 11$. Take half of $b$ (10) and square it: $(\frac{10}{2})^2 = 25$. Add 25 to both sides: $x^2 + 10x + 25 = 11 + 25$, which is $(x+5)^2 = 36$. So, $x+5 = \pm 6$, and $x=1$ or $x=-11$.

• Solve $x^2 - 8x - 9 = 0$. Move the constant: $x^2 - 8x = 9$. Add $(\frac{-8}{2})^2 = 16$ to both sides: $x^2 - 8x + 16 = 9+16$. This simplifies to $(x-4)^2 = 25$. So, $x-4 = \pm 5$, and $x=9$ or $x=-1$.

Property

Written in standard form, $ax^2 + bx + c = 0$, any quadratic equation can be solved using the quadratic formula:

where $a$, $b$, and $c$ are real numbers and $a \neq 0$. To use it, ensure the equation is in standard form, identify $a, b,$ and $c$, and substitute them into the formula.

Examples

• Solve $x^2 + 2x - 8 = 0$. Here $a=1, b=2, c=-8$. Using the formula: $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2}$. The solutions are $x=2$ and $x=-4$.

• Solve $3x^2 - 5x + 2 = 0$. Here $a=3, b=-5, c=2$. The formula gives $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(2)}}{2(3)} = \frac{5 \pm \sqrt{25-24}}{6} = \frac{5 \pm 1}{6}$. The solutions are $x=1$ and $x=\frac{2}{3}$.

Property

For $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real numbers, the discriminant is the expression under the radical in the quadratic formula: $b^2 - 4ac$. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

• If $b^2 - 4ac > 0$, there are two real solutions.
• If $b^2 - 4ac = 0$, there is one real solution.
• If $b^2 - 4ac < 0$, there are two complex solutions.

Examples

• For $x^2 - 10x + 25 = 0$, the discriminant is $(-10)^2 - 4(1)(25) = 100 - 100 = 0$. This indicates there is one rational solution.

• For $3x^2 + 7x - 6 = 0$, the discriminant is $7^2 - 4(3)(-6) = 49 + 72 = 121$. Since $121 > 0$ and is a perfect square, there are two rational solutions.