Quadratic Formula - Discriminant

RadfordMathematics.com

Online Mathematics Book

• Numbers
• Algebra
• Functions
• Geometry & Trigonometry
• Vectors & Matrices
• Statistics & Probabilities
• Calculus
• Search

×

Search Radford Mathematics

Close Quadratic Formula - Discriminant

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.

Quadratic Formula - Discriminant

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. \(6x^2-x-2=0\)

Answers w/out Working Answers with Working

Answers Without Working

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: \[x = \frac{2}{3} \quad \text{and} \quad x = -\frac{1}{2}\]

×

Answers with Working

Select the question number you'd like to see the working for:

Qu. 1 Qu. 2 Qu. 3 Qu. 4 Qu. 5 Close

Quadratics with Irrational Solutions

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. \(x^2 - 10x + 23 = 0\)

Answers w/out Working Answers with Working

Answers Without Working

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: \[x = 3.59 \quad \text{and} \quad x = 6.41\]