You are viewing the article How to Solve Quadratic Equations at Thptlaihoa.edu.vn you can quickly access the necessary information in the table of contents of the article below.
Quadratic equations, a fundamental concept in algebra, are encountered in various fields of science, engineering, and mathematics. These equations involve a variable raised to the power of two, which often creates complex mathematical expressions. Solving quadratic equations is a crucial skill that helps in determining the points where a quadratic function intersects the x-axis, finding the maximum or minimum values of a function, and understanding the behavior of parabolic curves. In this guide, we will explore various methods to solve quadratic equations, including factoring, using the quadratic formula, and completing the square. By understanding these techniques and implementing them strategically, you will gain the ability to find solutions to quadratic equations and gain valuable insights into the behavior of these mathematical functions. So, let’s dive into the world of quadratic equations and learn the art of solving them effectively.
This article is co-authored by a team of editors and trained researchers who confirm the accuracy and completeness of the article.
The wikiHow Content Management team carefully monitors the work of editors to ensure that every article is up to a high standard of quality.
This article has been viewed 89,620 times.
The first time you encounter a cubic equation (of the form ax3 + bx2 + cx + d = 0), you will probably feel more or less that the equation is unsolvable. However, the real way to solve cubic equations has been around for centuries! Discovered in the 1500s by Italian mathematicians Niccpò Tartaglia and Gerpamo Cardano, the way to solve cubic equations was one of the first not inherited from the ancient Greeks and Romans. While it can be quite difficult, with the right approach (and good background knowledge), even the most difficult cubic equations can be tamed.
Steps
Solve with quadratic formula
- On the other hand, if the equation contains constants, you will need another method of solving. Please refer to the ways to solve alternative equations below.
- Take for example the original cubic equation 3 x3 + -2 x2 + 14 x = 0. Taking an x as a common factor and moving it out of the equation, we get x (3 x2 + -2 x + 14) = 0 .
- In the above example, we will substitute the values of a, b and c (3, -2 and 14 respectively) into the quadratic equation as follows:
-
- {- b +/-√ ( b2 – 4 ac )}/2 a
- {-(-2) +/-√ ((-2) 2 – 4(3)(14))}/2(3)
- {2 +/-√ (4 – (12)(14))}/6
- {2 +/-√ (4 – (168)}/6
- {2 +/-√ (-164)}/6
-
- Experiment 1:
-
- {2 + (-164)}/6
- {2 + 12.8 i }/6
-
- Experiment 2:
-
- {2 – 12.8 i }/6
-
- The reason we can do this is because of the basic principle that any number multiplied by 0 equals 0 . When you break down the equation into the form x ( ax2 + bx + c ) = 0, you’ve essentially divided it into two “halves”: the half containing the variable x is on the left side, and the other half is the quadratic part lying on the left side. in brackets. If either of these “half” is zero, the whole equation will be zero. Thus, the two roots of the quadratic part are enclosed in brackets – two solutions will make this “half” zero, as will the zero itself – value that will make the left “half” zero, which is the solution of the cubic equation.
Find integer roots with a list of factors
- Take for example the equation 2 x3 + 9 x2 + 13 x = -6. In this case, to make the right side zero, we need to add both sides to 6. In the new equation, 2 x3 + 9 x2 + 13 x + 6 = 0, d = 6, therefore, cannot be applied. analysis tips above.
- In the example problem, a = 2 and d = 6 . The factors of 2 are 1 and 2 . The factors of 6 are 1, 2, 3 and 6 .
- In the above equation, taking the factor of a (1, 2) divided by the factor of d (1, 2, 3, 6), we get the following list: 1, 1/2, 1/3, 1/6 , 2 and 2/3. Next, we add negative values to complete the list: 1, -1, 1/2, -1/2, 1/3, -1/3, 1/6, -1/6, 2, – 2, 2/3, and -2/3 . The integer root of a cubic equation will be somewhere in this list.
- The Ruffini Rule is a complex subject. Here is an example of how to find a solution of a cubic equation with first degree polynomial division using the coefficients:
-
- -1 | 2 9 13 6
- __| -2-7-6
- __| 2 7 6 0
- Since the remainder is zero after all, we know that one of the integer roots of the equation is -1 .
-
Using the “differentiation” method
- For example, with the equation x3 – 3 x2 + 3 x – 1, we would write a = 1, b = -3, c = 3 and d = -1. Don’t forget that when there is no coefficient, it is quite possible to conclude that the variable x has a coefficient of 1.
- For the example problem, we do the following:
-
- b2 – 3 ac
- (-3) 2 – 3(1)(3)
- 9 – 3(1)(3)
- 9 – 9 = 0 = Δ0
-
- For the example problem, we do the following:
-
- 2(-3) 3 – 9(1)(-3)(3) + 27(1) 2 (-1)
- 2(-27) – 9(-9) + 27(-1)
- -54 + 81 – 27
- 81 – 81 = 0 = 1
-
- In the example problem, since both Δ0 and Δ1 = 0, finding Δ would be easy. We just do the following:
-
- 1 2 – 4Δ0 3 ) -27 a2
- (0) 2 – 4(0) 3 ) -27(1) 2
- 0 – 0 27
- 0 = Δ, so the equation has 1 or 2 solutions.
-
- In the example problem, we find C as follows:
-
- 3 (√((Δ1 2 – 4Δ0 3 ) + 1)/ 2)
- 3 (√((0 2 – 4(0) 3 ) + (0))/2)
- 3 (√((0 – 0) + (0))/2)
- 0 = C
-
- In the example, we could solve by testing the answers when n equals 1, 2, and 3. The answers obtained from these tests are possible solutions of a cubic equation — any value. give the result of 0 when substituting the equation will be the correct solution of the equation. For example, if 1 is obtained from one of the attempts, by substituting 1 in x3 – 3 x2 + 3 x – 1 will result in 0, so 1 is one of the solutions to the equation given third degree.
This article is co-authored by a team of editors and trained researchers who confirm the accuracy and completeness of the article.
The wikiHow Content Management team carefully monitors the work of editors to ensure that every article is up to a high standard of quality.
This article has been viewed 89,620 times.
The first time you encounter a cubic equation (of the form ax3 + bx2 + cx + d = 0), you will probably feel more or less that the equation is unsolvable. However, the real way to solve cubic equations has been around for centuries! Discovered in the 1500s by Italian mathematicians Niccpò Tartaglia and Gerpamo Cardano, the way to solve cubic equations was one of the first not inherited from the ancient Greeks and Romans. While it can be quite difficult, with the right approach (and good background knowledge), even the most difficult cubic equations can be tamed.
In conclusion, quadratic equations can be solved using a variety of methods, including factoring, completing the square, and using the quadratic formula. These different techniques provide flexibility in determining the roots or solutions of quadratic equations. It is important to understand the nature of the equation and select the most appropriate method for solving. Practicing and gaining familiarity with each method can improve problem-solving skills and make solving quadratic equations more efficient and accurate. By understanding and applying these techniques, individuals can confidently solve quadratic equations and effectively tackle real-world problems that involve quadratic relationships.
Thank you for reading this post How to Solve Quadratic Equations at Thptlaihoa.edu.vn You can comment, see more related articles below and hope to help you with interesting information.
Related Search:
1. Methods for solving quadratic equations
2. Quadratic formula explained
3. Factoring quadratic equations
4. Completing the square in quadratic equations
5. Using the discriminant to solve quadratic equations
6. Step-by-step guide to solving quadratic equations
7. Different types of solutions in quadratic equations
8. Real and complex roots in quadratic equations
9. Applications of quadratic equations in real life
10. Common mistakes to avoid when solving quadratic equations