Lesson Objectives
• Demonstrate an understanding of how to Solve a Quadratic Equation by Completing the Square
• Demonstrate an understanding of how to write a Quadratic Equation in Standard Form: ax2 + bx + c = 0
• Learn how to derive the Quadratic Formula
• Learn about the possible solutions using the Discriminant: b2 - 4ac
• Learn how to Solve a Quadratic Equation with Imaginary Solutions using the Quadratic Formula

So far, we have learned how to solve quadratic equations by factoring and by completing the square. Factoring is fast and easy but does not work for all quadratic equations. Completing the square works for all quadratic equations, but the process is tedious and not efficient. In this section, we will introduce the quadratic formula. This formula will allow us to quickly obtain the solution for any quadratic equation. To obtain this formula, we begin with the general or standard form of a quadratic equation. Essentially, we will solve this equation for x, by completing the square. $$ax^2 + bx + c=0$$ Since we need the coefficient for x2 to be 1, we begin by dividing each side of the equation by a, the coefficient of x2. $$\require{cancel}\frac{\cancel{a}}{\cancel{a}}x^2 + \frac{b}{a}x + \frac{c}{a}=0$$ For our next step, we want to move the constant term c/a to the other side. Let's subtract c/a from both sides of the equation. $$x^2 + \frac{b}{a}x=-\frac{c}{a}$$ For our next step, we want to complete the square. This means we are changing the left side of the equation into a perfect square trinomial. In other words, we will have a trinomial on the left side of the equation that can be factored into a binomial squared. This will allow us to use our square root property and solve the equation. To complete the square, we add one-half of the coefficient of the first-degree term squared to both sides of the equation. $$x^2 + \frac{b}{a}x + \left(\frac{1}{2}\cdot \frac{b}{a}\right)^2=-\frac{c}{a}+ \left(\frac{1}{2}\cdot \frac{b}{a}\right)^2$$ $$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}=-\frac{c}{a}+ \frac{b^2}{4a^2}$$ The left side of our equation is a perfect square trinomial. We can use our special factoring formulas to factor the left side into a binomial squared. $$\left(x + \frac{b}{2a}\right)^2=-\frac{c}{a}+ \frac{b^2}{4a^2}$$ Let's get a common denominator on the right side. We can multiply the fraction -c/a by 4a/4a. $$\left(x + \frac{b}{2a}\right)^2=-\frac{4ac}{4a^2}+ \frac{b^2}{4a^2}$$ Now, we can write the left side with one denominator. $$\left(x + \frac{b}{2a}\right)^2=\frac{b^2 - 4ac}{4a^2}$$ For our next step, we will use the square root property. $$\sqrt{\left(x + \frac{b}{2a}\right)^2}=\pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$ $$x + \frac{b}{2a}=\pm \frac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}}$$ $$x + \frac{b}{2a}=\pm \frac{\sqrt{b^2 - 4ac}}{2a}$$ To solve for x, let's move the b/2a to the other side of the equation. $$x=- \frac{b}{2a}\pm \frac{\sqrt{b^2 - 4ac}}{2a}$$ Since we have a common denominator (2a), we can write the quadratic formula as: $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ The quadratic formula might seem like gibberish at first, but notice how we have x isolated on the left and on the right we see a, b, and c. When a quadratic equation is in standard form, we have a as the coefficient of the squared variable, b as the coefficient for the variable raised to the first power, and c as the constant. We can just plug in these values and obtain our solution. Let's look at a few examples.
Example 1: Solve each equation using the quadratic formula. $$-3x^2 + x - 3=-5x^2$$ Let's begin by writing our equation in standard form. We can do this by adding 5x2 to both sides of the equation. $$2x^2 + x - 3=0$$ Now, let's record our values for a, the coefficient of x2, b, the coefficient of x to the first power, and c, the constant term.
a = 2
b = 1
c = -3
Now that we have these values, we can plug into our quadratic formula. $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-1 \pm \sqrt{(1)^2 - 4(2)(-3)}}{2(2)}$$ $$x=\frac{-1 \pm \sqrt{25}}{4}$$ $$x=\frac{-1 \pm 5}{4}$$ We will have two solutions: $$x=\frac{-1 + 5}{4}$$ $$x=\frac{4}{4}$$ $$x=1$$ $$x=\frac{-1 - 5}{4}$$ $$x=\frac{-6}{4}$$ $$x=\frac{-3}{2}$$ $$x=1,-\frac{3}{2}$$ Example 2: Solve each equation using the quadratic formula. $$2x^2=-2x + 9$$ Let's begin by writing our equation in standard form. We can do this by adding 2x and -9 to both sides of the equation. $$2x^2 + 2x - 9=0$$ Now, let's record our values for a, the coefficient of x2, b, the coefficient of x to the first power, and c, the constant term.
a = 2
b = 2
c = -9
Now that we have these values, we can plug into our quadratic formula. $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-2 \pm \sqrt{(2)^2 - 4(2)(-9)}}{2(2)}$$ $$x=\frac{-2 \pm 2\sqrt{19}}{4}$$ Before we go any further, we can factor a 2 out from the numerator and cancel this with a factor of 2 in the denominator. $$x=\frac{\cancel{2}(-1 \pm \sqrt{19})}{\cancel{2}\cdot 2}$$ Now we obtain our two solutions: $$x=\frac{-1 \pm \sqrt{19}}{2}$$

### Using the Quadratic Formula with Imaginary Solutions

At this point, we should know how to deal with imaginary solutions, which are solutions that involve the imaginary unit i. Since i, the imaginary unit is defined as the square root of -1, we can state the following: $$\sqrt{-b}=i\sqrt{b}$$ These types of solutions are also referred to as nonreal complex solutions, since they can be written as a + bi. Let's use this fact to solve a quadratic equation with an imaginary solution.
Example 3: Solve each equation using the quadratic formula. $$15x^2 + 8x + 4=5x^2$$ Let's begin by writing our equation in standard form. We can do this by adding -5x2 to both sides of the equation. $$10x^2 + 8x + 4=0$$ Now, let's record our values for a, the coefficient of x2, b, the coefficient of x to the first power, and c, the constant term.
a = 10
b = 8
c = 4
Now that we have these values, we can plug into our quadratic formula. $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-8 \pm \sqrt{(8)^2 - 4(10)(4)}}{2(10)}$$ $$x=\frac{-8 \pm \sqrt{-96}}{20}$$ Since we have the square root of a negative number, we can use the imaginary unit i: $$x=\frac{-8 \pm i\sqrt{96}}{20}$$ $$x=\frac{-8 \pm 4i\sqrt{6}}{20}$$ Before we go any further, we can factor a 4 out from the numerator and cancel this with a factor of 4 in the denominator. $$x=\frac{\cancel{4}(-2 \pm i\sqrt{6})}{\cancel{4}\cdot 5}$$ Now we obtain our two solutions: $$x=\frac{-2 \pm i\sqrt{6}}{5}$$

### The Discriminant

When we work with the quadratic equation, the part under the square root symbol is known as the discriminant. $$b^2 - 4ac$$ The discriminant tells us how many solutions and what type of solutions we will obtain.
Scenario 1: $$b^2 - 4ac > 0$$ In this case, there are two real solutions.
Scenario 2: $$b^2 - 4ac=0$$ In this case, there is exactly one real solution. The reason for this is very simple. The plus or minus part outside of the square root symbol doesn't matter. We will have plus or minus zero, which is just zero.
Scenario 3: $$b^2 - 4ac < 0$$ In this case, we will have two imaginary solutions (nonreal complex solutions) since we will have the square root of a negative number. Let's look at some examples.
Example 4: Determine the number of solutions for the quadratic equation using the discriminant. $$7x^2 - 6x - 8=2 + 3x$$ First, we write our equation in standard form. $$7x^2 - 9x - 10=0$$ Now, we record our values for a, b, and c.
a = 7, b = -9, c = -10
Let's plug into the discriminant: $$b^2 - 4ac$$ $$(-9)^2 - 4(7)(-10)=361$$ Since the discriminant is greater than 0, we know we have two real solutions.
Example 5: Determine the number of solutions for the quadratic equation using the discriminant. $$-2x^2 + 11x - 25=8x - 11$$ First, we write our equation in standard form. $$-2x^2 + 3x - 14=0$$ Now, we record our values for a, b, and c.
a = -2, b = 3, c = -14
Let's plug into the discriminant: $$b^2 - 4ac$$ $$(3)^2 - 4(-2)(-14)=-103$$ Since the discriminant is less than 0, we know we have two imaginary solutions (nonreal complex solutions).

#### Skills Check:

Example #1

Solve each equation. $$-3x^2=-15 + 4x$$

A
$$x=-1, \frac{7}{3}$$
B
$$x=-3, \frac{5}{3}$$
C
$$x=-5, \frac{1}{5}$$
D
$$x=1, \frac{9}{5}$$
E
$$x=3, \frac{22}{3}$$

Example #2

Solve each equation. $$2x^2 + 5x=75$$

A
$$x=-\frac{15}{2}, 5$$
B
$$x=-\frac{17}{3}, 6$$
C
$$x=-\frac{11}{2}, 9$$
D
$$x=-1, 4$$
E
$$x=-2, 9$$

Example #3

Solve each equation. $$-12x^2=-6x + 2$$

A
$$x=-1, \frac{1}{4}$$
B
$$x=\frac{-1 \pm i\sqrt{3}}{7}$$
C
$$x=\frac{-2 \pm i\sqrt{19}}{12}$$
D
$$x=\frac{3 \pm i\sqrt{15}}{12}$$
E
$$x=-\frac{9}{7}, 5$$