Factor And Solve Quadratic Equations A Step-by-Step Guide
Unlocking quadratic equations through factoring is a fundamental skill in algebra, forming the bedrock for more advanced mathematical concepts. Activity 4, aptly titled "Factor Then Solve!", provides a practical approach to mastering this skill. This comprehensive guide will delve into the solutions of the given quadratic equations, offering step-by-step explanations and insights to enhance understanding. Factoring, in essence, is the process of breaking down a quadratic expression into its constituent linear factors. Solving quadratic equations by factoring relies on the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This principle allows us to transform a quadratic equation into a pair of simpler linear equations, which can then be solved independently. The ability to solve quadratic equations is not merely an academic exercise; it has widespread applications in various fields, including physics, engineering, economics, and computer science. For example, quadratic equations can be used to model projectile motion, design suspension bridges, optimize financial investments, and develop algorithms for computer graphics. Thus, mastering factoring techniques is a valuable asset for anyone pursuing a career in a STEM field or any discipline that relies on mathematical modeling.
Problem 1: x² + 7x = 0
Solving quadratic equations starts with identifying common factors. In the equation x² + 7x = 0, we observe that 'x' is a common factor in both terms. Factoring out 'x', we get:
x(x + 7) = 0
Now, applying the zero-product property, we set each factor equal to zero:
x = 0 or x + 7 = 0
Solving these linear equations, we find the solutions:
x = 0 or x = -7
Therefore, the solutions to the quadratic equation x² + 7x = 0 are x = 0 and x = -7. These solutions represent the points where the parabola defined by the equation intersects the x-axis. In graphical terms, the roots of the equation correspond to the x-intercepts of the parabola. Understanding this connection between algebraic solutions and graphical representations is crucial for developing a deeper understanding of quadratic equations. Moreover, the process of factoring out common factors is a fundamental technique that extends to more complex factoring problems, making it an essential skill to master.
Problem 2: 6s² + 18s = 0
Factoring quadratic equations often involves finding the greatest common factor (GCF). In the equation 6s² + 18s = 0, the GCF of the coefficients 6 and 18 is 6, and 's' is a common variable factor. Factoring out 6s, we get:
6s(s + 3) = 0
Applying the zero-product property, we set each factor equal to zero:
6s = 0 or s + 3 = 0
Solving these linear equations, we find the solutions:
s = 0 or s = -3
Thus, the solutions to the quadratic equation 6s² + 18s = 0 are s = 0 and s = -3. The presence of a coefficient in front of the squared term does not fundamentally change the factoring process; we simply need to ensure that we factor out the GCF correctly. This problem highlights the importance of careful observation and attention to detail when factoring quadratic equations. Furthermore, the ability to identify and factor out GCFs is a crucial skill that simplifies the factoring process and reduces the likelihood of errors.
Problem 3: x² + 8x + 16 = 0
Perfect square trinomials provide a shortcut to factoring quadratic equations. The equation x² + 8x + 16 = 0 is a perfect square trinomial because it can be expressed as the square of a binomial. Recognizing this pattern simplifies the factoring process. We can rewrite the equation as:
(x + 4)² = 0
This is because (x + 4)² = (x + 4)(x + 4) = x² + 4x + 4x + 16 = x² + 8x + 16. Taking the square root of both sides, we get:
x + 4 = 0
Solving for x, we find the solution:
x = -4
In this case, we have a repeated root, meaning that the quadratic equation has only one distinct solution. This corresponds to the vertex of the parabola touching the x-axis at a single point. Recognizing perfect square trinomials is a valuable skill that can save time and effort when factoring quadratic equations. It also provides insights into the nature of the solutions and the graphical representation of the equation.
Problem 4: x² - 10x + 25 = 0
Another perfect square trinomial demonstrates the elegance of factoring patterns. The equation x² - 10x + 25 = 0 is also a perfect square trinomial, but with a negative middle term. We can rewrite the equation as:
(x - 5)² = 0
This is because (x - 5)² = (x - 5)(x - 5) = x² - 5x - 5x + 25 = x² - 10x + 25. Taking the square root of both sides, we get:
x - 5 = 0
Solving for x, we find the solution:
x = 5
Similar to Problem 3, this equation has a repeated root, indicating that the parabola touches the x-axis at only one point. The ability to recognize and factor perfect square trinomials with both positive and negative middle terms is essential for efficient problem-solving in algebra. It also reinforces the understanding of algebraic patterns and their connection to graphical representations.
Problem 5: h² + 6h = 16
Setting quadratic equations to zero is a crucial step before factoring. The equation h² + 6h = 16 is not in the standard form of a quadratic equation (ax² + bx + c = 0). To solve it by factoring, we first need to subtract 16 from both sides:
h² + 6h - 16 = 0
Now, we need to find two numbers that multiply to -16 and add up to 6. These numbers are 8 and -2. Therefore, we can factor the quadratic expression as:
(h + 8)(h - 2) = 0
Applying the zero-product property, we set each factor equal to zero:
h + 8 = 0 or h - 2 = 0
Solving these linear equations, we find the solutions:
h = -8 or h = 2
Thus, the solutions to the quadratic equation h² + 6h = 16 are h = -8 and h = 2. This problem highlights the importance of rearranging the equation into standard form before attempting to factor. It also reinforces the process of finding the correct factors by considering the product and sum of the roots.
Problem 6: r² - 14 = 5r
Rearranging and factoring: a common approach to solving quadratic equations. The equation r² - 14 = 5r is another example of a quadratic equation that needs to be rearranged before factoring. Subtracting 5r from both sides, we get:
r² - 5r - 14 = 0
Now, we need to find two numbers that multiply to -14 and add up to -5. These numbers are -7 and 2. Therefore, we can factor the quadratic expression as:
(r - 7)(r + 2) = 0
Applying the zero-product property, we set each factor equal to zero:
r - 7 = 0 or r + 2 = 0
Solving these linear equations, we find the solutions:
r = 7 or r = -2
Thus, the solutions to the quadratic equation r² - 14 = 5r are r = 7 and r = -2. This problem further emphasizes the importance of rearranging the equation into standard form and carefully considering the signs when finding the factors.
Problem 7: 11r + 15 = -2r²
Dealing with coefficients in front of the squared term requires careful factoring. The equation 11r + 15 = -2r² needs to be rearranged into standard form. Adding 2r² to both sides, we get:
2r² + 11r + 15 = 0
Now, we need to factor the quadratic expression. This involves finding two binomials that multiply to give the quadratic expression. We can use the "ac method" or trial and error. In this case, the factored form is:
(2r + 5)(r + 3) = 0
Applying the zero-product property, we set each factor equal to zero:
2r + 5 = 0 or r + 3 = 0
Solving these linear equations, we find the solutions:
r = -5/2 or r = -3
Thus, the solutions to the quadratic equation 11r + 15 = -2r² are r = -5/2 and r = -3. This problem demonstrates the factoring process when the coefficient of the squared term is not 1, which often requires more steps and careful consideration.
Problem 8: x² - 25 = 0
Difference of squares pattern: a powerful tool for factoring. The equation x² - 25 = 0 is a classic example of the difference of squares pattern, which states that a² - b² = (a + b)(a - b). Recognizing this pattern allows for quick and easy factoring. We can rewrite the equation as:
(x + 5)(x - 5) = 0
Applying the zero-product property, we set each factor equal to zero:
x + 5 = 0 or x - 5 = 0
Solving these linear equations, we find the solutions:
x = -5 or x = 5
Thus, the solutions to the quadratic equation x² - 25 = 0 are x = -5 and x = 5. The difference of squares pattern is a fundamental algebraic identity that simplifies factoring and is widely applicable in various mathematical contexts.
Problem 9: 81 - x² = 0
Applying the difference of squares pattern with a reversed order. The equation 81 - x² = 0 is another example of the difference of squares pattern, but with the terms in a reversed order. We can rewrite the equation as:
(9 + x)(9 - x) = 0
Applying the zero-product property, we set each factor equal to zero:
9 + x = 0 or 9 - x = 0
Solving these linear equations, we find the solutions:
x = -9 or x = 9
Thus, the solutions to the quadratic equation 81 - x² = 0 are x = -9 and x = 9. This problem reinforces the importance of recognizing the difference of squares pattern, even when the terms are not in the standard order. It also highlights the flexibility and power of algebraic identities in simplifying problem-solving.
Problem 10: Discussion on Quadratic Equations
Understanding the nature of quadratic equations through discussion. Quadratic equations are polynomial equations of the second degree, meaning that the highest power of the variable is 2. They have the general form ax² + bx + c = 0, where a, b, and c are constants and a ≠0. The solutions to a quadratic equation are also called roots or zeros, and they represent the x-intercepts of the parabola defined by the equation. A quadratic equation can have two distinct real roots, one repeated real root, or two complex roots. The nature of the roots is determined by the discriminant, which is given by the formula Δ = b² - 4ac. If Δ > 0, the equation has two distinct real roots. If Δ = 0, the equation has one repeated real root. If Δ < 0, the equation has two complex roots. Factoring is just one method for solving quadratic equations. Other methods include completing the square, using the quadratic formula, and graphical methods. Each method has its advantages and disadvantages, and the choice of method often depends on the specific equation and the desired level of accuracy. The study of quadratic equations is a cornerstone of algebra, and their applications extend to various fields, including physics, engineering, economics, and computer science. A deep understanding of quadratic equations is essential for success in higher-level mathematics and related disciplines. The discussion on quadratic equations should encompass various aspects, including the different methods for solving them, the nature of their roots, and their applications in real-world scenarios. This comprehensive approach will foster a deeper understanding of quadratic equations and their significance in mathematics and beyond. Furthermore, exploring the connections between algebraic solutions and graphical representations can provide valuable insights into the behavior of quadratic functions and their applications.
