Find The Second Solution To Quadratic Equations

Quadratic equations are a fundamental part of algebra, and understanding their solutions is key to mastering the subject. Typically, a quadratic equation, in the form of $ax^2 + bx + c = 0$ where $a$ is not equal to 0, has two solutions. These solutions, also known as roots, can be real or complex numbers. Today, we're going to dive into a fun challenge: given one solution to four different quadratic equations, we need to find the second solution for each. This exercise is particularly insightful when dealing with complex solutions, as they often come in conjugate pairs.

The Conjugate Pairs Theorem: A Key to Unlocking Solutions

One of the most important theorems when working with quadratic equations with real coefficients is the Conjugate Pairs Theorem. This theorem states that if a polynomial equation with real coefficients has a complex number $a + bi$ as a root, then its complex conjugate, $a - bi$, must also be a root. This is a powerful tool because it means that if we're given one complex solution, we can immediately deduce the other, provided the coefficients of the quadratic equation are real numbers (which is usually the case in standard problems unless otherwise specified). Think of it as a rule of nature for these types of equations: complex roots always come in pairs!

Let's break down how this applies to our specific problems. For each given solution, we'll assume the quadratic equation has real coefficients. This assumption is crucial. If the coefficients were allowed to be complex, then the conjugate pair theorem wouldn't necessarily hold, and the second solution could be anything. However, in most introductory and intermediate algebra contexts, problems are designed with real coefficients to leverage this theorem. So, when you see a complex number like $5 + 4i$ as a solution, you can confidently look for its twin, $5 - 4i$, to be the other solution.

This principle extends beyond just finding the second root. Understanding conjugate pairs is also vital when factoring polynomials, constructing polynomials from given roots, and even in more advanced topics like signal processing and control theory where complex numbers are used to model oscillating systems. The elegance of the conjugate pairs theorem lies in its simplicity and broad applicability within the realm of real-coefficient polynomials. It's a cornerstone concept that simplifies the process of finding all roots, especially when dealing with the intricacies of complex numbers. So, next time you encounter a complex root, remember its conjugate partner is likely waiting to be discovered!

Unpacking the First Quadratic Equation: $x = 5 + 4i$

Let's tackle the first equation, where we are given one solution as $x = 5 + 4i$. To find the second solution, we'll apply the Conjugate Pairs Theorem. This theorem tells us that if a quadratic equation has real coefficients, and $a + bi$ is a root, then its complex conjugate $a - bi$ must also be a root. In our case, the given solution is $5 + 4i$. Here, $a = 5$ and $b = 4$. The complex conjugate of $5 + 4i$ is obtained by simply changing the sign of the imaginary part. Therefore, the complex conjugate is $5 - 4i$. So, the second solution for the quadratic equation that has $x = 5 + 4i$ as one of its roots is $x = 5 - 4i$.

To illustrate why this works, consider a quadratic equation with roots $r_1$ and $r_2$. The equation can be written as $(x - r_1)(x - r_2) = 0$. If $r_1 = 5 + 4i$ and $r_2 = 5 - 4i$, let's expand this:

$(x - (5 + 4i))(x - (5 - 4i)) = 0$

We can rearrange the terms to group the real and imaginary parts:

$((x - 5) - 4i)((x - 5) + 4i) = 0$

This is in the form $(A - B)(A + B) = A^2 - B^2$, where $A = (x - 5)$ and $B = 4i$. So, we have:

$(x - 5)^2 - (4i)^2 = 0$

Now, let's expand and simplify:

$(x^2 - 10x + 25) - (16i^2) = 0$

Since $i^2 = -1$, we substitute that in:

$(x^2 - 10x + 25) - (16(-1)) = 0$

$x^2 - 10x + 25 + 16 = 0$

$x^2 - 10x + 41 = 0$

As you can see, the resulting quadratic equation $x^2 - 10x + 41 = 0$ has real coefficients (1, -10, and 41). This confirms that our application of the Conjugate Pairs Theorem was valid for this scenario. The theorem is indeed a reliable shortcut for finding the second complex root when the coefficients are real. This first example neatly demonstrates the power and direct applicability of the theorem.

Exploring the Second Quadratic Equation: $x = -4 - 5i$

Moving on to our second quadratic equation, we are given a single solution: $x = -4 - 5i$. Just as before, we assume this quadratic equation possesses real coefficients. This assumption is the bedrock upon which we build our solution. The Conjugate Pairs Theorem is our guiding principle here. It mandates that if a polynomial with real coefficients has a complex number $a + bi$ as a root, then its conjugate, $a - bi$, must also be a root. For the given root $x = -4 - 5i$, we can identify $a = -4$ and $b = -5$. To find its complex conjugate, we negate the imaginary part. Therefore, the complex conjugate of $-4 - 5i$ is $-4 - (-5i)$, which simplifies to $x = -4 + 5i$. This is our second solution.

Let's perform a quick verification, similar to the first case. If the roots are $r_1 = -4 - 5i$ and $r_2 = -4 + 5i$, the quadratic equation can be expressed as $(x - r_1)(x - r_2) = 0$. Substituting our roots:

$(x - (-4 - 5i))(x - (-4 + 5i)) = 0$

Rearranging to group real and imaginary components:

$((x + 4) + 5i)((x + 4) - 5i) = 0$

This again fits the $(A + B)(A - B) = A^2 - B^2$ pattern, with $A = (x + 4)$ and $B = 5i$. Applying the formula:

$(x + 4)^2 - (5i)^2 = 0$

Expanding and simplifying:

$(x^2 + 8x + 16) - (25i^2) = 0$

Substituting $i^2 = -1$:

$(x^2 + 8x + 16) - (25(-1)) = 0$

$x^2 + 8x + 16 + 25 = 0$

$x^2 + 8x + 41 = 0$

The resulting quadratic equation, $x^2 + 8x + 41 = 0$, clearly demonstrates that all its coefficients (1, 8, and 41) are real numbers. This consistency reinforces the validity of using the Conjugate Pairs Theorem. It's a reliable method for uncovering the hidden root when you're presented with one complex root and the implicit understanding of real coefficients. This exercise in finding the second root for $x = -4 - 5i$ further solidifies the theorem's practical utility.

Analyzing the Third Quadratic Equation: $x = 4 + 5i$

Now, let's analyze the third quadratic equation. We are given one of its solutions as $x = 4 + 5i$. As with the previous examples, our foundational assumption is that this quadratic equation is characterized by real coefficients. This assumption is absolutely critical for applying the Conjugate Pairs Theorem. The theorem states that for a polynomial equation with real coefficients, if $a + bi$ is a root, then its complex conjugate $a - bi$ must also be a root. For the provided solution $x = 4 + 5i$, we identify $a = 4$ and $b = 5$. To find the complex conjugate, we simply change the sign of the imaginary part. Thus, the complex conjugate of $4 + 5i$ is $4 - 5i$. Therefore, the second solution for this quadratic equation is $x = 4 - 5i$.

Let's verify this by constructing the quadratic equation using both roots, $r_1 = 4 + 5i$ and $r_2 = 4 - 5i$. The equation can be written as $(x - r_1)(x - r_2) = 0$. Substituting our roots:

$(x - (4 + 5i))(x - (4 - 5i)) = 0$

Rearranging to group the real and imaginary parts:

$((x - 4) - 5i)((x - 4) + 5i) = 0$

This fits the $(A - B)(A + B) = A^2 - B^2$ pattern, where $A = (x - 4)$ and $B = 5i$. Applying the formula:

$(x - 4)^2 - (5i)^2 = 0$

Expanding and simplifying:

$(x^2 - 8x + 16) - (25i^2) = 0$

Since $i^2 = -1$, we substitute:

$(x^2 - 8x + 16) - (25(-1)) = 0$

$x^2 - 8x + 16 + 25 = 0$

$x^2 - 8x + 41 = 0$

The resulting quadratic equation, $x^2 - 8x + 41 = 0$, has coefficients (1, -8, and 41) that are all real numbers. This verification confirms that our application of the Conjugate Pairs Theorem is sound. It's a reliable and elegant method for finding the second root when dealing with complex solutions and implicitly real coefficients. This exploration for $x = 4 + 5i$ further underscores the theorem's importance in simplifying algebraic problems involving complex numbers.

Deconstructing the Fourth Quadratic Equation: $x = 5 - 4i$

Finally, we arrive at the fourth quadratic equation, where one solution is provided as $x = 5 - 4i$. Our steadfast assumption remains that this quadratic equation is defined by real coefficients. This condition is paramount for the correct application of the Conjugate Pairs Theorem. The theorem assures us that if a polynomial equation with real coefficients has a root in the form of $a + bi$, then its complex conjugate $a - bi$ must also be a root. For the given solution $x = 5 - 4i$, we can identify $a = 5$ and $b = -4$. To find the complex conjugate, we invert the sign of the imaginary component. Therefore, the complex conjugate of $5 - 4i$ is $5 - (-4i)$, which simplifies to $x = 5 + 4i$. This is the second solution.

Let's confirm this by constructing the quadratic equation with roots $r_1 = 5 - 4i$ and $r_2 = 5 + 4i$. The equation form is $(x - r_1)(x - r_2) = 0$. Plugging in our roots:

$(x - (5 - 4i))(x - (5 + 4i)) = 0$

Rearranging to group real and imaginary parts:

$((x - 5) + 4i)((x - 5) - 4i) = 0$

This again follows the $(A + B)(A - B) = A^2 - B^2$ pattern, with $A = (x - 5)$ and $B = 4i$. Applying the formula:

$(x - 5)^2 - (4i)^2 = 0$

Expanding and simplifying:

$(x^2 - 10x + 25) - (16i^2) = 0$

Substituting $i^2 = -1$:

$(x^2 - 10x + 25) - (16(-1)) = 0$

$x^2 - 10x + 25 + 16 = 0$

$x^2 - 10x + 41 = 0$

The resulting quadratic equation, $x^2 - 10x + 41 = 0$, has all real coefficients (1, -10, and 41). This consistent outcome validates our use of the Conjugate Pairs Theorem. It's a reliable shortcut for finding the complementary complex root when the coefficients of the quadratic equation are real. This final example reinforces the theorem's efficacy in solving such problems.

Conclusion: The Power of Conjugate Pairs

In summary, for each of the four quadratic equations presented, assuming they have real coefficients, the second solution is the complex conjugate of the given solution. This is a direct application of the Conjugate Pairs Theorem.

• For $x = 5 + 4i$, the second solution is $x = 5 - 4i$.
• For $x = -4 - 5i$, the second solution is $x = -4 + 5i$.
• For $x = 4 + 5i$, the second solution is $x = 4 - 5i$.
• For $x = 5 - 4i$, the second solution is $x = 5 + 4i$.

This exercise highlights a fundamental property of quadratic equations with real coefficients: their complex roots always appear in conjugate pairs. Understanding and applying this theorem can significantly simplify the process of finding all solutions. It's a testament to the structured and predictable nature of mathematics, especially within the realm of algebra.

For further exploration into the fascinating world of quadratic equations and complex numbers, you might find the resources at Khan Academy to be incredibly helpful. They offer comprehensive explanations and practice problems that can deepen your understanding.