Lesson 4: Using the Quadratic Formula

Property

To derive the quadratic formula, start with the general form $ax^2 + bx + c = 0$ and complete the square:

1. Divide by $a$: $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
2. Move the constant: $x^2 + \frac{b}{a}x = -\frac{c}{a}$
3. Complete the square: $x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$
4. Factor and simplify: $(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$
5. Extract roots: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

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.