Solving $3(x^2-49)(36x^2-49)=0$ A Step-by-Step Guide

Introduction to Solving Polynomial Equations

In the realm of mathematics, particularly in algebra, solving polynomial equations is a fundamental skill. Polynomial equations are expressions that involve variables raised to non-negative integer powers, combined with coefficients and constants. Finding the solutions, also known as roots or zeros, of a polynomial equation means identifying the values of the variable that make the equation true. These solutions are critical in various fields, including engineering, physics, economics, and computer science. Understanding how to solve polynomial equations is essential for modeling real-world phenomena and making accurate predictions.

One common technique for solving polynomial equations is the factoring method. Factoring involves breaking down a complex polynomial expression into simpler factors, often linear or quadratic expressions. Once the polynomial is factored, the zero-product property can be applied. This property states that if the product of several factors is zero, then at least one of the factors must be zero. By setting each factor equal to zero and solving for the variable, we can find the roots of the original polynomial equation. This method is particularly effective for equations that can be factored easily, such as those involving differences of squares or perfect square trinomials. However, not all polynomial equations can be factored easily, and other methods, such as the quadratic formula or numerical techniques, may be required.

For higher-degree polynomial equations, various techniques can be employed. Synthetic division is a streamlined method for dividing a polynomial by a linear factor, which can help in identifying roots and factoring the polynomial further. The Rational Root Theorem provides a systematic way to find potential rational roots of a polynomial equation. It states that if a polynomial equation has integer coefficients, then any rational root must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. By testing these potential rational roots, we can often find at least one root, which can then be used to reduce the degree of the polynomial and simplify the problem. In cases where finding exact solutions is difficult or impossible, numerical methods such as the Newton-Raphson method or the bisection method can be used to approximate the roots to a desired level of accuracy. These methods involve iterative processes that converge towards the solutions, providing valuable insights into the behavior of the polynomial equation.

Detailed Analysis of the Equation $3(x^2-49)(36x^2-49)=0$

The given equation is $3(x^2-49)(36x^2-49)=0$. To solve this equation, we'll use the zero-product property, which states that if the product of several factors is zero, then at least one of the factors must be zero. This means we need to set each factor equal to zero and solve for $x$. The factors in this equation are $3$, $(x^2-49)$, and $(36x^2-49)$.

The first factor is the constant $3$. Setting this equal to zero gives us $3 = 0$, which is not possible. Therefore, this factor does not contribute any solutions to the equation. The second factor is $(x^2-49)$. This is a difference of squares, which can be factored as $(x-7)(x+7)$. Setting this factor equal to zero gives us $(x-7)(x+7) = 0$. Applying the zero-product property again, we set each sub-factor equal to zero: $x-7 = 0$ and $x+7 = 0$. Solving these equations gives us $x = 7$ and $x = -7$. These are two solutions to the original equation.

The third factor is $(36x^2-49)$. This is also a difference of squares, and it can be factored as $(6x-7)(6x+7)$. Setting this factor equal to zero gives us $(6x-7)(6x+7) = 0$. Applying the zero-product property again, we set each sub-factor equal to zero: $6x-7 = 0$ and $6x+7 = 0$. Solving these equations gives us $6x = 7$ and $6x = -7$. Dividing by 6, we get x = rac{7}{6} and x = -rac{7}{6}. These are two more solutions to the original equation. In summary, by factoring the given equation and applying the zero-product property, we have found four solutions: $x = 7$, $x = -7$, x = rac{7}{6}, and x = -rac{7}{6}. These values of $x$ make the original equation true, and they are the roots of the polynomial equation.

Step-by-Step Solution Using Factoring and the Zero-Product Property

To solve the equation $3(x^2-49)(36x^2-49)=0$, we will follow a step-by-step approach using factoring and the zero-product property. This method is particularly effective for polynomial equations that can be easily factored, allowing us to find the solutions systematically.

Step 1: Identify the Factors The equation is given in a factored form: $3(x^2-49)(36x^2-49)=0$. We can see three primary factors: the constant $3$, the quadratic expression $(x^2-49)$, and the quadratic expression $(36x^2-49)$. The goal is to find the values of $x$ that make the entire expression equal to zero. This involves setting each factor equal to zero and solving for $x$.

Step 2: Apply the Zero-Product Property The zero-product property states that if the product of several factors is zero, then at least one of the factors must be zero. Mathematically, if $A imes B imes C = 0$, then either $A = 0$, $B = 0$, or $C = 0$ (or any combination thereof). Applying this to our equation, we set each factor equal to zero:

1. $3 = 0$ (This is not possible, so it does not yield a solution.)
2. $x^2 - 49 = 0$
3. $36x^2 - 49 = 0$

Step 3: Factor the Quadratic Expressions Both $(x^2-49)$ and $(36x^2-49)$ are differences of squares, which can be factored using the formula $a^2 - b^2 = (a - b)(a + b)$. For the first quadratic expression, $(x^2-49)$, we have $a = x$ and $b = 7$, so it factors to $(x-7)(x+7)$. For the second quadratic expression, $(36x^2-49)$, we have $a = 6x$ and $b = 7$, so it factors to $(6x-7)(6x+7)$. Now, our equations become:

1. $(x-7)(x+7) = 0$
2. $(6x-7)(6x+7) = 0$

Step 4: Solve for $x$ Applying the zero-product property again to each factored equation:

For $(x-7)(x+7) = 0$, we have:

1. $x - 7 = 0 ightarrow x = 7$
2. $x + 7 = 0 ightarrow x = -7$

For $(6x-7)(6x+7) = 0$, we have:

1. 6x - 7 = 0 ightarrow 6x = 7 ightarrow x = rac{7}{6}
2. 6x + 7 = 0 ightarrow 6x = -7 ightarrow x = -rac{7}{6}

Step 5: List the Solutions The solutions to the equation $3(x^2-49)(36x^2-49)=0$ are:

• $x = 7$
• $x = -7$
• x = rac{7}{6}
• x = -rac{7}{6}

These are the four values of $x$ that make the original equation true. By using the factoring method and the zero-product property, we have systematically found all the solutions.

Verification of Solutions

To ensure the accuracy of our solutions, it is crucial to verify each value in the original equation $3(x^2-49)(36x^2-49)=0$. This process confirms that the solutions we found are indeed the roots of the polynomial equation. We will substitute each solution back into the equation and check if the result is zero.

Verification for $x = 7$ Substitute $x = 7$ into the equation:

$3((7)^2-49)(36(7)^2-49) = 3(49-49)(36(49)-49) = 3(0)(1764-49) = 3(0)(1715) = 0$ The equation holds true for $x = 7$, so it is a valid solution.

Verification for $x = -7$ Substitute $x = -7$ into the equation:

$3((-7)^2-49)(36(-7)^2-49) = 3(49-49)(36(49)-49) = 3(0)(1764-49) = 3(0)(1715) = 0$ The equation holds true for $x = -7$, confirming it as a valid solution.

Verification for x = rac{7}{6} Substitute x = rac{7}{6} into the equation:

3igg(ig(rac{7}{6}ig)^2-49igg)igg(36ig(rac{7}{6}ig)^2-49igg) = 3igg(rac{49}{36}-49igg)igg(36ig(rac{49}{36}ig)-49igg)

= 3igg(rac{49}{36}-49igg)(49-49) = 3igg(rac{49}{36}-49igg)(0) = 0 The equation holds true for x = rac{7}{6}, so it is a valid solution.

Verification for x = -rac{7}{6} Substitute x = -rac{7}{6} into the equation:

3igg(ig(-rac{7}{6}ig)^2-49igg)igg(36ig(-rac{7}{6}ig)^2-49igg) = 3igg(rac{49}{36}-49igg)igg(36ig(rac{49}{36}ig)-49igg)

= 3igg(rac{49}{36}-49igg)(49-49) = 3igg(rac{49}{36}-49igg)(0) = 0 The equation holds true for x = -rac{7}{6}, confirming it as a valid solution.

By verifying each solution in the original equation, we have confirmed that $x = 7$, $x = -7$, x = rac{7}{6}, and x = -rac{7}{6} are indeed the roots of the equation $3(x^2-49)(36x^2-49)=0$. This step is essential to ensure the accuracy of our solutions and to demonstrate a comprehensive understanding of the problem-solving process.

Conclusion

In conclusion, solving the equation $3(x^2-49)(36x^2-49)=0$ involves applying the zero-product property and factoring techniques. The given equation can be efficiently solved by recognizing the factors and setting each of them equal to zero. The key steps include factoring the quadratic expressions $(x^2-49)$ and $(36x^2-49)$ into $(x-7)(x+7)$ and $(6x-7)(6x+7)$, respectively, and then applying the zero-product property to find the values of $x$ that satisfy the equation. This method provides a systematic way to determine the roots of the polynomial equation.

We identified four distinct solutions: $x = 7$, $x = -7$, x = rac{7}{6}, and x = -rac{7}{6}. Each of these values makes the original equation true, as verified by substituting them back into the equation. The solutions $x = 7$ and $x = -7$ arise from the factor $(x^2-49)$, while the solutions x = rac{7}{6} and x = -rac{7}{6} come from the factor $(36x^2-49)$. The constant factor of $3$ does not contribute any solutions, as it cannot be equal to zero.

Understanding how to solve polynomial equations through factoring and the zero-product property is a crucial skill in algebra. This approach is not only applicable to this specific equation but also serves as a foundation for solving more complex polynomial equations. Furthermore, the ability to factor and apply the zero-product property is essential in various mathematical contexts and real-world applications, including calculus, physics, and engineering. The detailed analysis and step-by-step solution provided here offer a clear understanding of the process and reinforce the importance of these fundamental algebraic techniques.