Solving Quadratic Equations A Comprehensive Guide
In the realm of algebra, quadratic equations stand as fundamental building blocks. They appear in various scientific and engineering applications, making their comprehension essential. This article delves into the process of solving quadratic equations, interpreting their roots, and exploring their diverse applications. We'll dissect three distinct quadratic equation problems, offering step-by-step solutions and insightful explanations. Whether you're a student grappling with algebra or a professional seeking a refresher, this guide aims to equip you with a solid understanding of quadratic equations and their solutions.
Problem 1 Finding the Smaller Root of a Quadratic Equation
The first problem presents us with the quadratic equation 3x² + 14x - 5 = 0. Our goal is to identify the smaller root among the given options. To achieve this, we can employ several methods, including factoring, completing the square, or utilizing the quadratic formula.
Applying the Quadratic Formula
The quadratic formula is a versatile tool for solving any quadratic equation in the form ax² + bx + c = 0. It states that the roots are given by:
x = (-b ± √(b² - 4ac)) / (2a)
In our case, a = 3, b = 14, and c = -5. Substituting these values into the quadratic formula, we get:
x = (-14 ± √(14² - 4 * 3 * -5)) / (2 * 3) x = (-14 ± √(196 + 60)) / 6 x = (-14 ± √256) / 6 x = (-14 ± 16) / 6
This yields two possible solutions:
x₁ = (-14 + 16) / 6 = 2 / 6 = 1/3 x₂ = (-14 - 16) / 6 = -30 / 6 = -5
Comparing the two roots, 1/3 and -5, we find that the smaller root is -5. Therefore, the correct answer is option 2) -5.
Factoring the Quadratic Equation
Alternatively, we can attempt to factor the quadratic equation. Factoring involves expressing the quadratic expression as a product of two binomials. For 3x² + 14x - 5 = 0, we seek two numbers that multiply to (3 * -5) = -15 and add up to 14. These numbers are 15 and -1.
We can rewrite the middle term (14x) as 15x - x, leading to:
3x² + 15x - x - 5 = 0
Now, we factor by grouping:
3x(x + 5) - 1(x + 5) = 0 (3x - 1)(x + 5) = 0
Setting each factor to zero gives us the roots:
3x - 1 = 0 => x = 1/3 x + 5 = 0 => x = -5
As before, we identify -5 as the smaller root, confirming our previous result.
Importance of Choosing the Right Method
While both the quadratic formula and factoring can solve quadratic equations, the choice of method often depends on the specific equation. The quadratic formula is universally applicable, while factoring is more efficient when the equation can be easily factored. In this case, factoring required a bit of insight, whereas the quadratic formula was a straightforward application of a formula. Understanding the strengths and weaknesses of each method allows for efficient problem-solving.
Problem 2 Finding the Modulus of the Difference of Roots
The second problem asks us to find the modulus (absolute value) of the difference between the roots of the equation 2x² + 7x - 4 = 0. This problem not only requires us to find the roots but also to perform an additional step of calculating the modulus of their difference.
Using the Quadratic Formula Again
We once again employ the quadratic formula, with a = 2, b = 7, and c = -4:
x = (-7 ± √(7² - 4 * 2 * -4)) / (2 * 2) x = (-7 ± √(49 + 32)) / 4 x = (-7 ± √81) / 4 x = (-7 ± 9) / 4
This gives us the two roots:
x₁ = (-7 + 9) / 4 = 2 / 4 = 1/2 x₂ = (-7 - 9) / 4 = -16 / 4 = -4
Calculating the Modulus of the Difference
Now, we find the difference between the roots and then take its absolute value:
|x₁ - x₂| = |1/2 - (-4)| |x₁ - x₂| = |1/2 + 4| |x₁ - x₂| = |9/2| |x₁ - x₂| = 4.5
Therefore, the modulus of the difference of the roots is 4.5, which corresponds to option 4).
Alternative Approach Factoring and Verification
To reinforce our understanding, let's also solve this by factoring. We need two numbers that multiply to (2 * -4) = -8 and add up to 7. These numbers are 8 and -1.
2x² + 8x - x - 4 = 0 2x(x + 4) - 1(x + 4) = 0 (2x - 1)(x + 4) = 0
Setting each factor to zero gives us:
2x - 1 = 0 => x = 1/2 x + 4 = 0 => x = -4
These are the same roots we found using the quadratic formula. The modulus of their difference remains:
|1/2 - (-4)| = 4.5
This exercise highlights the consistency between different solution methods and reinforces the correctness of our answer. Choosing to verify solutions through alternative methods is a crucial practice in mathematical problem-solving.
Applications of Root Differences
The concept of the difference between roots has applications in various areas, such as analyzing the stability of systems in engineering or determining the range of parameters for which a system exhibits certain behaviors. Understanding how to calculate and interpret root differences provides a foundation for tackling more advanced problems.
Problem 3 Finding the Sum of Roots with a Squared Binomial
The third problem presents an equation involving a squared binomial: (2x - 1)² + 8x = 9. This equation requires us to first expand and simplify before we can apply our methods for finding roots. The goal is to find the sum of the roots of this equation.
Expanding and Simplifying the Equation
First, expand the squared binomial:
(2x - 1)² = (2x - 1)(2x - 1) = 4x² - 4x + 1
Now, substitute this back into the original equation:
4x² - 4x + 1 + 8x = 9
Combine like terms and rearrange to get the standard quadratic form:
4x² + 4x + 1 - 9 = 0 4x² + 4x - 8 = 0
We can simplify this equation by dividing all terms by 4:
x² + x - 2 = 0
Vieta's Formulas A Shortcut for Finding the Sum of Roots
While we could use the quadratic formula or factoring, there's a more efficient approach for finding the sum of the roots: Vieta's formulas. Vieta's formulas provide a direct relationship between the coefficients of a polynomial and the sums and products of its roots. For a quadratic equation ax² + bx + c = 0, Vieta's formulas state:
Sum of roots: x₁ + x₂ = -b/a Product of roots: x₁ * x₂ = c/a
In our simplified equation, x² + x - 2 = 0, we have a = 1, b = 1, and c = -2. Therefore, the sum of the roots is:
x₁ + x₂ = -1/1 = -1
Thus, the sum of the roots is -1.
Factoring to Verify Roots and their Sum
To verify our result, let's factor the simplified quadratic equation:
x² + x - 2 = 0 (x + 2)(x - 1) = 0
This gives us the roots:
x₁ = -2 x₂ = 1
The sum of these roots is -2 + 1 = -1, which confirms our answer obtained using Vieta's formulas.
Advantages of Using Vieta's Formulas
Vieta's formulas offer a significant advantage when the problem only requires the sum or product of roots, without needing the individual root values. This method saves time and effort compared to finding the roots explicitly and then summing them. Mastering Vieta's formulas is a valuable skill for solving quadratic equation problems efficiently.
Conclusion Mastering Quadratic Equations
In this article, we've explored three distinct problems involving quadratic equations. We've utilized the quadratic formula, factoring, and Vieta's formulas to find roots, their differences, and their sums. Each method offers unique advantages, and the choice of method often depends on the specific problem and the desired outcome.
Understanding quadratic equations is crucial for success in algebra and related fields. By mastering the techniques presented here, you'll be well-equipped to tackle a wide range of problems involving quadratic expressions. The quadratic formula, factoring, and Vieta's formulas are powerful tools in your mathematical arsenal. Remember to practice and apply these methods to solidify your understanding and enhance your problem-solving skills. This comprehensive approach to solving quadratic equations will undoubtedly benefit you in your future mathematical endeavors.
