Solving Quadratic Equations A Comprehensive Guide
In this comprehensive guide, we will delve into the fascinating world of quadratic equations, exploring various methods to solve them. Quadratic equations, characterized by their highest power of 2, are fundamental in mathematics and have numerous applications in diverse fields such as physics, engineering, and economics. We will tackle four distinct quadratic equations, each presenting its unique challenges and opportunities for applying different solution techniques. Our journey will cover factoring, the quadratic formula, and completing the square, equipping you with a versatile toolkit for conquering these equations.
a. Solving $10000x^2 - 64 = 0$ by Factoring and the Square Root Property
In this section, we embark on the first quadratic equation, $10000x^2 - 64 = 0$, which presents an excellent opportunity to demonstrate the power of factoring and the square root property. Factoring, a cornerstone of algebraic manipulation, involves expressing a quadratic equation as a product of two linear expressions. The square root property, on the other hand, leverages the fundamental relationship between squaring and taking the square root to isolate the variable. By mastering these techniques, you'll gain invaluable skills for tackling a wide range of quadratic equations.
To begin, we recognize that the given equation, $10000x^2 - 64 = 0$, is a difference of squares. This specific form allows us to apply a well-established factoring pattern: $a^2 - b^2 = (a + b)(a - b)$. Identifying this pattern is crucial, as it simplifies the equation significantly and paves the way for a straightforward solution. In our case, we can rewrite $10000x^2$ as $(100x)^2$ and $64$ as $8^2$. This transformation allows us to express the equation as the difference of two perfect squares, making the factoring process more intuitive.
Applying the difference of squares pattern, we factor the equation as follows: $(100x + 8)(100x - 8) = 0$. This factorization step is pivotal, as it transforms the quadratic equation into a product of two linear factors. The zero-product property then comes into play, stating that if the product of two factors is zero, at least one of the factors must be zero. This principle allows us to split the original equation into two simpler equations, each involving a single linear factor.
Setting each factor to zero, we obtain two linear equations: $100x + 8 = 0$ and $100x - 8 = 0$. These equations are significantly easier to solve than the original quadratic equation. To solve for $x$ in the first equation, we subtract 8 from both sides, yielding $100x = -8$. Dividing both sides by 100, we find $x = -8/100$, which simplifies to $x = -2/25$. Similarly, for the second equation, we add 8 to both sides, resulting in $100x = 8$. Dividing both sides by 100, we obtain $x = 8/100$, which simplifies to $x = 2/25$.
Alternatively, we can solve the equation using the square root property. This method involves isolating the squared term and then taking the square root of both sides. Starting with the original equation, $10000x^2 - 64 = 0$, we add 64 to both sides, giving us $10000x^2 = 64$. Next, we divide both sides by 10000, resulting in $x^2 = 64/10000$. Taking the square root of both sides, we must remember to consider both positive and negative roots, as squaring either a positive or negative number yields a positive result. Thus, we have $x = ±√(64/10000)$.
Simplifying the square root, we find $x = ±(8/100)$, which further simplifies to $x = ±(2/25)$. This result aligns perfectly with the solutions obtained through factoring, demonstrating the consistency and reliability of both methods. The solutions, $x = 2/25$ and $x = -2/25$, represent the points where the parabola defined by the quadratic equation intersects the x-axis. These points are also known as the roots or zeros of the quadratic equation.
In conclusion, the equation $10000x^2 - 64 = 0$ can be efficiently solved using both factoring and the square root property. Factoring leverages the difference of squares pattern, while the square root property involves isolating the squared term and taking the square root of both sides. Both methods yield the same solutions, $x = 2/25$ and $x = -2/25$, reinforcing the versatility of algebraic techniques in solving quadratic equations. Understanding and mastering these methods are crucial for building a strong foundation in algebra and its applications.
b. Solving $9x^2 - 8 = -34x$ by Factoring and the Quadratic Formula
Let's tackle the second quadratic equation, $9x^2 - 8 = -34x$, which presents a slightly different challenge. This equation requires us to rearrange the terms before applying our solution methods. We will explore two primary techniques: factoring and the quadratic formula. Factoring, as we've seen, involves expressing the quadratic as a product of linear factors. The quadratic formula, a powerful and versatile tool, provides a direct solution for any quadratic equation, regardless of its factorability. By mastering both approaches, you'll be well-equipped to handle a broader range of quadratic equations.
First, we need to rewrite the equation in the standard quadratic form, which is $ax^2 + bx + c = 0$. To achieve this, we add $34x$ to both sides of the equation $9x^2 - 8 = -34x$, resulting in $9x^2 + 34x - 8 = 0$. This rearrangement is crucial, as it sets the stage for both factoring and the quadratic formula. The standard form allows us to clearly identify the coefficients $a$, $b$, and $c$, which are essential for applying these methods.
Now, let's attempt to solve the equation by factoring. Factoring involves finding two binomials that, when multiplied, yield the quadratic expression. This process can sometimes be challenging, especially when the coefficients are not straightforward. In our case, we need to find two numbers that multiply to $(9)(-8) = -72$ and add up to $34$. This requires careful consideration of the factors of -72 and their sums. After some deliberation, we find that the numbers 36 and -2 satisfy these conditions, as $(36)(-2) = -72$ and $36 + (-2) = 34$.
Using these numbers, we can rewrite the middle term, $34x$, as $36x - 2x$. This technique, known as factoring by grouping, allows us to break down the quadratic expression into a form that is easier to factor. Substituting this into the equation, we get $9x^2 + 36x - 2x - 8 = 0$. Now, we group the first two terms and the last two terms: $(9x^2 + 36x) + (-2x - 8) = 0$. Factoring out the greatest common factor (GCF) from each group, we have $9x(x + 4) - 2(x + 4) = 0$.
Notice that we now have a common factor of $(x + 4)$ in both terms. Factoring this out, we obtain $(x + 4)(9x - 2) = 0$. This factorization step is the culmination of our efforts, transforming the quadratic equation into a product of two linear factors. Applying the zero-product property, we set each factor to zero: $x + 4 = 0$ and $9x - 2 = 0$.
Solving the first equation, $x + 4 = 0$, we subtract 4 from both sides, yielding $x = -4$. For the second equation, $9x - 2 = 0$, we add 2 to both sides, giving us $9x = 2$. Dividing both sides by 9, we find $x = 2/9$. Thus, the solutions obtained by factoring are $x = -4$ and $x = 2/9$.
Now, let's solve the same equation using the quadratic formula. The quadratic formula is a general solution for any quadratic equation of the form $ax^2 + bx + c = 0$, and it is given by: $x = (-b ± √(b^2 - 4ac)) / (2a)$. In our equation, $9x^2 + 34x - 8 = 0$, we have $a = 9$, $b = 34$, and $c = -8$. Substituting these values into the quadratic formula, we get:
Simplifying the expression under the square root, we have:
This gives us two possible solutions:
These solutions, $x = 2/9$ and $x = -4$, are identical to those obtained by factoring, demonstrating the consistency of the two methods. The quadratic formula provides a reliable alternative, especially when factoring proves difficult or impossible. It ensures that we can always find the solutions to a quadratic equation, regardless of its complexity.
In summary, the equation $9x^2 - 8 = -34x$ can be effectively solved using both factoring and the quadratic formula. Factoring involves rearranging the equation, finding the appropriate factors, and applying the zero-product property. The quadratic formula provides a direct solution by substituting the coefficients into a well-established formula. Both methods yield the same solutions, $x = -4$ and $x = 2/9$, highlighting the power and versatility of algebraic techniques in solving quadratic equations. Understanding and mastering these methods are crucial for building a strong foundation in algebra and its applications.
c. Solving $2x^2 - 4x + 7 = 0$ Using the Quadratic Formula
For the third quadratic equation, $2x^2 - 4x + 7 = 0$, we encounter a scenario where factoring may not be the most straightforward approach. In such cases, the quadratic formula shines as a reliable and efficient method. The quadratic formula, as we've discussed, provides a direct solution for any quadratic equation in the standard form $ax^2 + bx + c = 0$. By applying this formula, we can determine the nature of the solutions, whether they are real or complex, and find their precise values. This section will focus on demonstrating the power and versatility of the quadratic formula in handling quadratic equations that may not be easily factored.
Recall that the quadratic formula is given by: $x = (-b ± √(b^2 - 4ac)) / (2a)$. In our equation, $2x^2 - 4x + 7 = 0$, we can identify the coefficients as $a = 2$, $b = -4$, and $c = 7$. These coefficients are the key ingredients for applying the quadratic formula. Substituting these values into the formula, we get:
This substitution step is crucial, as it sets the stage for simplifying the expression and finding the solutions. The careful substitution of the coefficients ensures that we are applying the formula correctly and accurately.
Now, let's simplify the expression step by step. First, we simplify the terms inside the square root:
Notice that we have a negative value under the square root. This indicates that the solutions will be complex numbers, involving the imaginary unit $i$, where $i = √(-1)$. This is a significant observation, as it tells us that the parabola defined by the quadratic equation does not intersect the x-axis. The complex solutions represent points in the complex plane, rather than points on the real number line.
To further simplify the expression, we can rewrite $√(-40)$ as $√(4 * -10)$, which is equal to $2√(-10)$. Since $√(-1) = i$, we can write $2√(-10)$ as $2i√10$. Substituting this back into the equation, we have:
Now, we can simplify the expression by dividing both the real and imaginary parts by 4:
This gives us two complex solutions:
These solutions are complex conjugates, meaning they have the same real part but opposite imaginary parts. Complex solutions often arise in quadratic equations where the discriminant, $b^2 - 4ac$, is negative. The discriminant provides valuable information about the nature of the solutions, indicating whether they are real and distinct, real and equal, or complex conjugates.
In summary, the equation $2x^2 - 4x + 7 = 0$ has two complex solutions, which we found using the quadratic formula. The negative value under the square root indicated the presence of complex solutions. By simplifying the expression and using the imaginary unit $i$, we arrived at the solutions $x = 1 + (i√10) / 2$ and $x = 1 - (i√10) / 2$. This example highlights the power of the quadratic formula in handling equations that may not be easily factored and in revealing the nature of the solutions, whether they are real or complex. Understanding complex numbers and their role in quadratic equations is crucial for a comprehensive understanding of algebra and its applications.
d. Solving $3.2x + 0.2x^2 - 5 = 0$ Using the Quadratic Formula and Decimal Handling
Our final quadratic equation, $3.2x + 0.2x^2 - 5 = 0$, presents a unique challenge due to the presence of decimal coefficients. While the core method for solving this equation remains the quadratic formula, we'll need to carefully handle the decimals to ensure accuracy and clarity. This section will not only demonstrate the application of the quadratic formula but also provide insights into working with decimals in algebraic equations. By mastering these techniques, you'll be able to confidently solve quadratic equations with decimal coefficients and appreciate the importance of precision in mathematical calculations.
First, let's rewrite the equation in the standard quadratic form, $ax^2 + bx + c = 0$. Rearranging the terms in $3.2x + 0.2x^2 - 5 = 0$, we get $0.2x^2 + 3.2x - 5 = 0$. This rearrangement is crucial for identifying the coefficients $a$, $b$, and $c$, which are essential for applying the quadratic formula. In this case, we have $a = 0.2$, $b = 3.2$, and $c = -5$.
Before applying the quadratic formula, it's often beneficial to eliminate the decimals to simplify the calculations. We can do this by multiplying the entire equation by a suitable power of 10. In this case, multiplying by 10 will clear the decimals, giving us $2x^2 + 32x - 50 = 0$. This transformation maintains the equality of the equation while making the coefficients integers, which can simplify the subsequent calculations. However, for demonstration purposes and to showcase the handling of decimals, we will proceed with the original decimal coefficients.
Now, let's apply the quadratic formula: $x = (-b ± √(b^2 - 4ac)) / (2a)$. Substituting the decimal coefficients, we get:
This substitution step is critical, as it sets the stage for simplifying the expression and finding the solutions. The careful substitution of the coefficients ensures that we are applying the formula correctly and accurately, even with decimal values.
Next, we simplify the expression step by step. First, we simplify the terms inside the square root:
Now, we need to find the square root of 14.24. Using a calculator or a square root approximation method, we find that $√14.24 ≈ 3.774$. Substituting this value back into the equation, we have:
This gives us two possible solutions:
Thus, the solutions to the equation $3.2x + 0.2x^2 - 5 = 0$ are approximately $x ≈ 1.435$ and $x ≈ -17.435$. These solutions represent the points where the parabola defined by the quadratic equation intersects the x-axis.
In summary, the equation $3.2x + 0.2x^2 - 5 = 0$ can be effectively solved using the quadratic formula, even with decimal coefficients. The key is to carefully substitute the coefficients and simplify the expression step by step. While eliminating decimals by multiplying the equation by a power of 10 can simplify calculations, it's also important to be comfortable working with decimals directly. The approximate solutions, $x ≈ 1.435$ and $x ≈ -17.435$, highlight the importance of precision in mathematical calculations and the power of the quadratic formula in solving a wide range of quadratic equations. This example demonstrates the versatility of the quadratic formula and its applicability in various scenarios, including those involving decimal coefficients.
Conclusion
In this comprehensive guide, we've explored various methods for solving quadratic equations, including factoring, the square root property, and the quadratic formula. We've tackled four distinct quadratic equations, each presenting its unique challenges and opportunities for applying different solution techniques. From equations that can be easily factored to those requiring the quadratic formula for complex solutions, we've covered a wide spectrum of quadratic equation scenarios. By mastering these methods, you'll gain a strong foundation in algebra and its applications, empowering you to confidently solve quadratic equations in various contexts. The ability to solve quadratic equations is a fundamental skill in mathematics, with applications spanning diverse fields such as physics, engineering, and economics. The techniques and insights gained in this guide will serve as valuable tools in your mathematical journey and beyond.