Finding The Value Of K In The Equation X^2 + Kx + 6 = (x + 2)(x + 3)

In the realm of algebra, quadratic equations hold a prominent position, serving as fundamental tools for modeling various real-world phenomena. Among the key aspects of these equations are the coefficients that govern their behavior. When faced with an equation like $x^2 + kx + 6 = (x + 2)(x + 3)$, our quest is to unravel the mystery surrounding the value of 'k'. This exploration will not only enhance our understanding of quadratic equations but also equip us with valuable problem-solving techniques applicable in diverse mathematical contexts.

Unveiling the Equation's Secrets

To embark on this mathematical journey, let's first dissect the given equation: $x^2 + kx + 6 = (x + 2)(x + 3)$. At first glance, we notice that it presents a quadratic expression on the left-hand side and a factored form on the right-hand side. Our mission is to determine the value of 'k' that makes this equation hold true for all possible values of 'x'. This implies that the two sides of the equation must be equivalent, representing the same quadratic function. To achieve this, we'll employ the powerful technique of expanding the factored form and comparing coefficients.

The right-hand side of the equation, $(x + 2)(x + 3)$, can be expanded using the distributive property, often referred to as the FOIL method (First, Outer, Inner, Last). This process involves multiplying each term in the first factor by each term in the second factor, resulting in:

$(x + 2)(x + 3) = x * x + x * 3 + 2 * x + 2 * 3$

Simplifying this expression, we get:

$x^2 + 3x + 2x + 6 = x^2 + 5x + 6$

Now, we have the expanded form of the right-hand side: $x^2 + 5x + 6$. Our original equation can now be rewritten as:

$x^2 + kx + 6 = x^2 + 5x + 6$

The Coefficient Connection

The next crucial step in our quest involves comparing the coefficients of the corresponding terms on both sides of the equation. The coefficients are the numerical values that multiply the variables. In this case, we have the coefficients of $x^2$, $x$, and the constant terms.

On the left-hand side, the coefficient of $x^2$ is 1, the coefficient of $x$ is 'k', and the constant term is 6. On the right-hand side, the coefficient of $x^2$ is also 1, the coefficient of $x$ is 5, and the constant term is 6.

For the equation to hold true for all values of 'x', the coefficients of the corresponding terms must be equal. This leads us to a crucial observation: the coefficient of $x$ on the left-hand side ('k') must be equal to the coefficient of $x$ on the right-hand side (5). Therefore, we can confidently conclude that:

$k = 5$

This elegant solution unveils the value of 'k' that satisfies the given equation. By expanding the factored form and comparing coefficients, we've successfully navigated the algebraic landscape and arrived at our destination.

The Significance of Coefficient Comparison

The technique of comparing coefficients is a cornerstone of algebra, providing a powerful tool for solving equations and establishing relationships between different expressions. It hinges on the fundamental principle that two polynomials are equal if and only if their corresponding coefficients are equal. This principle allows us to equate coefficients of like terms, transforming complex equations into simpler algebraic relationships that are easier to solve.

In the context of quadratic equations, comparing coefficients enables us to determine unknown parameters, such as the value of 'k' in our example. By equating the coefficients of $x$, we were able to directly solve for 'k', avoiding more convoluted methods like substituting specific values of 'x'. This technique shines in its efficiency and elegance, offering a streamlined approach to solving algebraic problems.

Beyond quadratic equations, coefficient comparison extends its reach to polynomials of higher degrees and systems of equations. It serves as a unifying thread in algebraic problem-solving, empowering us to tackle a wide range of mathematical challenges.

The Answer

Therefore, the value of k is 5, which corresponds to option (d).

Exploring the Broader Implications

The problem we've tackled here serves as a stepping stone to more advanced algebraic concepts. Understanding how to manipulate and solve quadratic equations is crucial for various mathematical disciplines, including calculus, linear algebra, and differential equations. Moreover, quadratic equations find applications in numerous real-world scenarios, such as modeling projectile motion, optimizing designs, and analyzing financial data.

By mastering the techniques involved in solving equations like $x^2 + kx + 6 = (x + 2)(x + 3)$, we're not only honing our algebraic skills but also equipping ourselves with tools that are essential for success in STEM fields and beyond. The ability to expand expressions, compare coefficients, and solve for unknowns is a fundamental asset in any problem-solving endeavor.

In conclusion, the journey to find the value of 'k' in the given equation has been an insightful exploration of quadratic equations and the power of coefficient comparison. This technique, grounded in the principles of algebra, empowers us to solve equations efficiently and elegantly. As we continue our mathematical pursuits, let's carry with us the lessons learned here, applying them to new challenges and expanding our understanding of the mathematical world.

Practice Problems

To solidify your understanding of this concept, try solving these practice problems:

1. If $x^2 + px + 10 = (x + 2)(x + 5)$, find the value of p.
2. If $x^2 - qx + 12 = (x - 3)(x - 4)$, find the value of q.
3. If $2x^2 + rx + 3 = (2x + 1)(x + 3)$, find the value of r.

These problems offer an opportunity to apply the techniques we've discussed and further develop your algebraic problem-solving skills. Remember, practice is key to mastering any mathematical concept.

By diligently working through these exercises, you'll gain confidence in your ability to tackle quadratic equations and similar algebraic challenges. Embrace the process of learning, and you'll find that the world of mathematics is filled with exciting discoveries and rewarding achievements.

Find the value of k if $x^2 + kx + 6 = (x + 2)(x + 3)$ holds true.

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