### Online Mathematics Book

In this section we learn how to solve quadratic equations.
By the end of this section, we'll know how to solve equations like: $6x^2-x-2 = 0$

We start by watching the tutorial, below. We then read through a summary of the two-step method for solving quadratic equations that we see in the tutorial. Finally we work through some questions each of which has both an answer key and video detailed solutions.

## Tutorial

The following tutorial covers all of the essentials we'll need to know about the quadratic formula.
It lasts a little more than $$10$$ minutes, do take notes! This should help you understand/know how to use the quadratic formula.

Given a quadratic equation: $ax^2+bx+c = 0$ we can solve this equation for $$x$$ using the following two steps:

• Step 1: calculate the discriminant, using the formula: $\Delta = b^2 - 4ac$
• Step 2: solve the quadratic equation, which depends on the sign of the discriminant $$\Delta$$, which leads to three possible cases:
• Case 1 if $$\Delta >0$$: then the quadratic equation has two solutions: $x = \frac{-b-\sqrt{\Delta}}{2a} \quad \text{and} \quad x = \frac{-b+\sqrt{\Delta}}{2a}$ we'll often write these two solutions, using the $$\pm$$ symbol as: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$
• Case 2 if $$\Delta = 0$$: then the quadratic equation has one solution: $x = \frac{-b}{2a}$ In this case the solution is often referred to as the double root.
• Case 3 if $$\Delta < 0$$: then the quadratic equation has no real solution.

## Exercise 1

Using the quadratic formula, solve each of the following for $$x$$:

1. $$2x^2+4x-6 = 0$$
2. $$-x^2+3x+10=0$$
3. $$x^2-x-12 = 0$$
4. $$2x^2+12x+18 = 0$$
5. $$3x^2+x-2=0$$
6. $$3x^2-6x+8 = 0$$
7. $$-x^2+10x-25 = 0$$
8. $$-5x^2+3x+2=0$$
9. $$x^2+4x+7=0$$
10. $$6x^2-x-2=0$$

1. The solution(s) to $$2x^2+4x-6 = 0$$ are: $x = 1 \quad \text{and} \quad x = -3$
2. The solution(s) to $$-x^2+3x+10=0$$ are: $x = -2 \quad \text{and} \quad x = 5$
3. The solution(s) to $$x^2-x-12 = 0$$ are: $x = -3 \quad \text{and} \quad x = 4$
4. The solution(s) to $$2x^2+12x+18 = 0$$ are: $x = -3$
5. The solution(s) to $$3x^2+x-2=0$$ are: $x = -1 \quad \text{and} \quad x = \frac{2}{3}$
6. The solution(s) to $$3x^2-6x+8 = 0$$ are: $\text{No Solutions!}$
7. The solution(s) to $$-x^2+10x-25 = 0$$ are: $x = 5$
8. The solutions to $$-5x^2+3x+2=0$$ are: $x = 1 \quad \text{and} \quad x = -\frac{2}{5}$
9. The solutions to $$x^2+4x+7=0$$ are: $\text{No Solutions!}$
10. The solutions to $$6x^2-x-2=0$$ are: $x = \frac{2}{3} \quad \text{and} \quad x = -\frac{1}{2}$

Solution(s) to quadratic equations won't always be "nice" round numbers.

For example, we'll be seeing below, the quadratic equation: $-3x^2+5x-1 = 0$ Has solutions: $x = \frac{5+\sqrt{13}}{6} \quad \text{and} \quad x = \frac{5-\sqrt{13}}{6}$ We'll be faced with this type of solution, in which the $$x$$ values are irrational numbers, as soon as the discriminant $$\Delta$$ isn't a square number (the square root therefore cannot be simplified).

## Exercise 2

Solve each of the following for $$x$$.
Write your answers in irrational form and (using your calculator) write their values rounded to three significant figures.

1. $$x^2-4x-1 = 0$$
2. $$2x^2 - 3x - 7 = 0$$
3. $$-3x^2+6x+3 = 0$$
4. $$-x^2+2x+2 = 0$$
5. $$x^2 - 10x + 23 = 0$$

1. The solution(s) to $$x^2-4x-1 = 0$$ is: $x = 2- \sqrt{5} \quad \text{and} \quad x = 2+\sqrt{5}$ Calculating these and rounding to 3 significant figures leads to: $x = -0.236 \quad \text{and} \quad x = 4.236$
2. The solution(s) to $$2x^2 - 3x - 7 = 0$$ is: $x = \frac{3-\sqrt{65}}{4} \quad \text{and} \quad x = \frac{3+\sqrt{65}}{4}$ Calculating these and rounding to 3 significant figures leads to: $x = -1.27 \quad \text{and} \quad x = 2.77$
3. The solution(s) to $$-3x^2+6x+3 = 0$$ is: $x = 1-\sqrt{2} \quad \text{and} \quad x = 1+\sqrt{2}$ Calculating these and rounding to 3 significant figures leads to: $x = -0.414 \quad \text{and} \quad x = 2.414$
4. The solution(s) to $$-x^2+2x+2 = 0$$ is: $x = 1-\sqrt{3} \quad \text{and} \quad x = 1+\sqrt{3}$ Calculating these and rounding to 3 significant figures leads to: $x = -0.732 \quad \text{and} \quad x = 2.73$
5. The solution(s) to $$x^2 - 10x + 23 = 0$$ is: $x = 5-\sqrt{2} \quad \text{and} \quad x = 5+\sqrt{2}$ Calculating these and rounding to 3 significant figures leads to: $x = 3.59 \quad \text{and} \quad x = 6.41$