Lesson 1: Quadratic Formula

New Concept

The Quadratic Formula provides a powerful, universal method for solving any quadratic equation of the form $ax^2 + bx + c = 0$. It gives the solutions directly from the coefficients, saving you from completing the square every time.

What’s next

Next, you'll master this formula with interactive examples and practice cards. We'll then apply it to real-world challenges and explore complex number solutions.

Property

The solutions of the equation $ax^2 + bx + c = 0$, $(a \neq 0)$ are

This formula provides two solutions, represented by the $\pm$ symbol:

Examples

• To solve $2x^2 + 1 = 4x$, first write it in standard form as $2x^2 - 4x + 1 = 0$. With $a=2$, $b=-4$, and $c=1$, the formula gives $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)} = \frac{4 \pm \sqrt{8}}{4}$.

• To solve $x^2 - 3x = 1$, rewrite it as $x^2 - 3x - 1 = 0$. With $a=1$, $b=-3$, and $c=-1$, the formula yields $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2}$.

Property

Imaginary Unit:
$i = \sqrt{-1}$ or $i^2 = -1$

Imaginary Numbers:
If $a \ge 0$, then $\sqrt{-a} = \sqrt{-1}\sqrt{a} = i\sqrt{a}$.

The sum of a real number and an imaginary number is called a complex number.

Property

The discriminant of a quadratic equation is

1. If $D > 0$, there are two unequal real solutions.
2. If $D = 0$, there is one solution of multiplicity two.
3. If $D < 0$, there are two complex conjugate solutions.

Examples

• For $y = x^2 - x - 3$, the discriminant is $D = (-1)^2 - 4(1)(-3) = 13$. Since $D > 0$, the equation has two distinct real solutions and the graph has two x-intercepts.

• For $y = 2x^2 + x + 1$, the discriminant is $D = 1^2 - 4(2)(1) = -7$. Since $D < 0$, the equation has two complex solutions and the graph has no x-intercepts.

Property

To solve a formula that is quadratic for one variable in terms of others, first write the equation in standard form $ax^2 + bx + c = 0$ with respect to that variable. The coefficients $a$, $b$, and $c$ may contain other variables. Then, apply the quadratic formula.

Examples

• To solve $x^2 - xy + y = 2$ for $x$, rearrange it to $x^2 - (y)x + (y - 2) = 0$. Using the formula with $a=1$, $b=-y$, and $c=y-2$ gives $x = \frac{y \pm \sqrt{y^2 - 4y + 8}}{2}$.

• To solve the height equation $h = 4t - 16t^2$ for time $t$, rewrite it as $16t^2 - 4t + h = 0$. The quadratic formula gives $t = \frac{4 \pm \sqrt{16 - 64h}}{32}$.