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Quadratic equations are a fundamental concept in the field of algebra, and their solutions play a crucial role in various disciplines, including physics, engineering, and finance. These equations take the form of ax² + bx + c = 0, where a, b, and c are constants, and x represents the variable we’re solving for. While the process of solving quadratic equations may seem daunting at first, it is actually quite straightforward once you understand the key techniques and formulas involved. In this guide, we will explore the stepbystep process of how to solve quadratic equations, providing clear explanations and examples along the way. Whether you are a student looking to improve your algebra skills or an individual seeking to apply quadratic equations in reallife scenarios, mastering this topic will undoubtedly enhance your problemsolving abilities and broaden your mathematical horizons.
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The first time you encounter a cubic equation (of the form ax ^{3} + bx ^{2} + 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 x ^{3} + 2 x ^{2} + 14 x = 0. Taking an x as a common factor and moving it out of the equation, we get x (3 x ^{2} + 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 +/√ ( b ^{2} – 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 ( ax ^{2} + 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 x ^{3} + 9 x ^{2} + 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 x ^{3} + 9 x ^{2} + 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
 __ 276
 __ 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 x ^{3} – 3 x ^{2} + 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:

 b ^{2} – 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 a ^{2}
 (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 x ^{3} – 3 x ^{2} + 3 x – 1 will result in 0, so 1 is one of the solutions to the equation given third degree.
This article is coauthored 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 ax ^{3} + bx ^{2} + 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, solving quadratic equations can be approached using various methods, such as factoring, completing the square, or applying the quadratic formula. Each method has its own advantages and limitations, depending on the nature of the equation. It is important to carefully analyze the given equation and choose the most suitable method accordingly. Additionally, practicing and becoming familiar with these solving techniques can greatly enhance one’s problemsolving skills in mathematics. Quadratic equations are widely applicable in various fields, and being able to solve them effectively will not only contribute to academic success but also provide a solid foundation for advanced mathematical concepts.
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