Solving $3x + \sqrt{x-1} = 6$: Finding The Quadratic Equation

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n mathematics, solving equations involving radicals often requires careful manipulation to eliminate the radical terms and transform the equation into a more manageable form, such as a quadratic equation. In this article, we will walk through the process of finding the correct quadratic equation to solve for the possible solutions of the equation $3x + \sqrt{x-1} = 6$. This involves isolating the radical, squaring both sides, and simplifying the resulting equation. Understanding these steps is crucial for anyone studying algebra and equation solving. Let's dive in and explore how to tackle this type of problem step by step.

Isolating the Radical Term

To begin, our main goal here is to isolate the square root term in the equation $3x + \sqrt{x-1} = 6$. This is a crucial first step because it sets us up to eliminate the radical by squaring both sides of the equation. By isolating the radical, we ensure that when we square both sides, the square root will be effectively removed, simplifying the equation into a more familiar form. This simplification is the key to solving equations with radicals.

The initial equation is:

$3x + \sqrt{x-1} = 6$

To isolate the square root, we need to subtract $3x$ from both sides of the equation. This gives us:

$\sqrt{x-1} = 6 - 3x$

Now, we have the square root term all by itself on one side of the equation, which is exactly what we want. This prepares us for the next step, where we will square both sides to eliminate the square root. Remember, the strategy here is to gradually transform the equation into a form that is easier to solve, and isolating the radical is a significant step in that direction. By following this approach, we can systematically solve even complex equations involving radicals. The importance of this step cannot be overstated, as it lays the foundation for the subsequent steps and ultimately leads us to the solution.

Squaring Both Sides

With the radical term isolated, the next step in solving the equation $3x + \sqrt{x-1} = 6$ is to square both sides. Squaring both sides eliminates the square root, which is essential for simplifying the equation and making it solvable. However, it's crucial to remember that squaring both sides can sometimes introduce extraneous solutions, so we'll need to check our answers later to ensure they are valid. This is a standard caveat when dealing with radical equations, and careful verification is always necessary.

We have the equation:

$\sqrt{x-1} = 6 - 3x$

Now, we square both sides:

$(\sqrt{x-1})^2 = (6 - 3x)^2$

This simplifies the left side to $x-1$, as the square root and the square cancel each other out. On the right side, we need to expand $(6 - 3x)^2$. Recall that $(a - b)^2 = a^2 - 2ab + b^2$. Applying this formula, we get:

$(6 - 3x)^2 = 6^2 - 2(6)(3x) + (3x)^2 = 36 - 36x + 9x^2$

So, our equation now looks like this:

$x - 1 = 9x^2 - 36x + 36$

This is a significant step forward because we've eliminated the radical and transformed the equation into a quadratic form, which we can then rearrange and solve using standard methods. Squaring both sides is a powerful technique, but it always comes with the responsibility of checking for extraneous solutions. The resulting quadratic equation is a critical stepping stone towards finding the possible values of $x$ that satisfy the original equation.

Rearranging into Quadratic Form

Now that we have eliminated the radical and obtained the equation $x - 1 = 9x^2 - 36x + 36$, our next critical step is to rearrange it into the standard quadratic form, which is $ax^2 + bx + c = 0$. This form allows us to easily identify the coefficients $a$, $b$, and $c$, which are necessary for applying various methods to solve quadratic equations, such as factoring, completing the square, or using the quadratic formula. The standard form provides a structured way to analyze the equation and determine the solutions.

Starting with:

$x - 1 = 9x^2 - 36x + 36$

We want to move all terms to one side of the equation, leaving zero on the other side. To do this, we can subtract $x$ and add $1$ to both sides. This gives us:

$0 = 9x^2 - 36x + 36 - x + 1$

Now, we combine like terms:

$0 = 9x^2 - 37x + 37$

This equation is now in the standard quadratic form $ax^2 + bx + c = 0$, where $a = 9$, $b = -37$, and $c = 37$. Having the equation in this form is a significant achievement because it allows us to directly apply methods for solving quadratic equations. The process of rearranging the equation into standard form is a fundamental skill in algebra and is essential for solving a wide range of problems. By achieving this, we are now well-positioned to find the solutions for $x$ that satisfy the quadratic equation.

Identifying the Correct Quadratic Equation

After rearranging the equation, we arrived at the quadratic form $9x^2 - 37x + 37 = 0$. Now, our objective is to identify which of the given options matches this equation. This step is straightforward but essential to ensure we have correctly performed the algebraic manipulations. Matching the derived equation with the provided options confirms our work and directs us to the correct answer.

We have the quadratic equation:

$9x^2 - 37x + 37 = 0$

Comparing this to the given options:

• A. $9x^2 - 37x + 37 = 0$
• B. $9x^2 - 35x + 37 = 0$
• C. $9x^2 - x + 37 = 0$
• D. $9x^2 + x + 37 = 0$

It is clear that option A, $9x^2 - 37x + 37 = 0$, exactly matches the quadratic equation we derived. Therefore, option A is the correct answer. This step highlights the importance of careful comparison and attention to detail when working through mathematical problems. Verifying the result against the given options ensures accuracy and confirms that the solution aligns with the problem's requirements. By correctly identifying the quadratic equation, we've successfully completed a significant part of the problem-solving process.

Checking for Extraneous Solutions

Having found the quadratic equation $9x^2 - 37x + 37 = 0$, it's now imperative to consider the possibility of extraneous solutions. Extraneous solutions are solutions that arise from the process of solving the equation (in this case, squaring both sides) but do not satisfy the original equation. They are a common issue when dealing with radical equations, and failing to check for them can lead to incorrect answers. The process of checking for extraneous solutions involves substituting the solutions back into the original equation to verify their validity.

The original equation was:

$3x + \sqrt{x-1} = 6$

We need to solve the quadratic equation $9x^2 - 37x + 37 = 0$ to find potential values for $x$. The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a = 9$, $b = -37$, and $c = 37$. Plugging these values into the quadratic formula, we get:

$x = \frac{37 \pm \sqrt{(-37)^2 - 4(9)(37)}}{2(9)}$

$x = \frac{37 \pm \sqrt{1369 - 1332}}{18}$

$x = \frac{37 \pm \sqrt{37}}{18}$

So, we have two potential solutions:

$x_1 = \frac{37 + \sqrt{37}}{18}$ and $x_2 = \frac{37 - \sqrt{37}}{18}$

Now, we must substitute each of these values back into the original equation $3x + \sqrt{x-1} = 6$ to check for extraneous solutions. This step involves careful arithmetic and algebraic manipulation. If a value of $x$ does not satisfy the original equation, it is an extraneous solution and must be discarded. The process of checking for extraneous solutions is a critical component of solving radical equations and ensures that only valid solutions are accepted. While the computations can be a bit involved, the accuracy and correctness of the final answer depend on it.

Conclusion

In summary, solving the equation $3x + \sqrt{x-1} = 6$ involved several key steps: isolating the radical term, squaring both sides, rearranging the equation into quadratic form, identifying the correct quadratic equation, and crucially, checking for extraneous solutions. We found that the correct quadratic equation to solve is $9x^2 - 37x + 37 = 0$. This process illustrates the importance of careful algebraic manipulation and the necessity of verifying solutions in radical equations. By following these steps, we can confidently solve complex equations and ensure the accuracy of our results.

For further information on solving radical equations and quadratic equations, you might find helpful resources on websites like Khan Academy, which offers comprehensive lessons and practice exercises.