Quadratic Polynomial With Zeros √2 And √3 A Step By Step Guide

In mathematics, particularly in algebra, quadratic polynomials play a fundamental role. A quadratic polynomial is a polynomial of degree two, and it can be expressed in the general form ${ax^2 + bx + c}$, where ${a}$, ${b}$, and ${c}$ are constants and ${a \neq 0}$. The zeros or roots of a quadratic polynomial are the values of ${x}$ for which the polynomial equals zero. Knowing the roots of a quadratic polynomial allows us to construct the polynomial itself.

This article delves into the process of finding a quadratic polynomial given its roots. Specifically, we will explore how to determine the quadratic polynomial that has ${\sqrt{2}}$ and ${\sqrt{3}}$ as its zeros. This involves understanding the relationship between the roots and the coefficients of a quadratic polynomial and applying the formulas derived from this relationship. We will also discuss common pitfalls and alternative methods for constructing quadratic polynomials, providing a comprehensive guide for students and enthusiasts of algebra.

Understanding the Relationship Between Roots and Coefficients

To effectively find a quadratic polynomial from its roots, it's crucial to understand the relationship between the roots and the coefficients of the polynomial. For a quadratic polynomial of the form ${ax^2 + bx + c = 0}$, let ${\alpha}$ and ${\beta}$ be the roots. According to Vieta's formulas, the sum and product of the roots are related to the coefficients as follows:

• Sum of the roots: ${\alpha + \beta = -\frac{b}{a}}$
• Product of the roots: ${\alpha \beta = \frac{c}{a}}$

These relationships are fundamental in constructing a quadratic polynomial when its roots are known. By using these formulas, we can determine the coefficients of the polynomial and thereby define the polynomial itself. This approach provides a direct method for solving problems where roots are given, and the corresponding quadratic polynomial is to be found.

Constructing the Quadratic Polynomial

Given the roots ${\alpha}$ and ${\beta}$, the quadratic polynomial can be constructed using the following general form:

${k(x^2 - (\alpha + \beta)x + \alpha \beta)}$

Here, ${k}$ is a non-zero constant. This constant scales the polynomial but does not change its roots. For simplicity, ${k}$ is often taken to be 1. Substituting ${k = 1}$, we get:

${x^2 - (\alpha + \beta)x + \alpha \beta}$

This form directly uses the sum and product of the roots to define the quadratic polynomial. It's a powerful and efficient way to construct a quadratic polynomial, especially when dealing with real or complex roots. Understanding this construction method is key to solving problems related to quadratic polynomials in algebra.

Finding the Quadratic Polynomial with Roots √2 and √3

Now, let's apply this knowledge to find the specific quadratic polynomial with roots ${\sqrt{2}}$ and ${\sqrt{3}}$. Here, ${\alpha = \sqrt{2}}$ and ${\beta = \sqrt{3}}$. We will follow the steps outlined above to construct the polynomial.

Step 1: Calculate the Sum of the Roots

The sum of the roots, denoted as ${\alpha + \beta}$, is calculated as:

${\sqrt{2} + \sqrt{3}}$

This is a straightforward addition of two irrational numbers. The sum ${\sqrt{2} + \sqrt{3}}$ will be used in the linear term of the quadratic polynomial.

Step 2: Calculate the Product of the Roots

The product of the roots, denoted as ${\alpha \beta}$, is calculated as:

${\sqrt{2} \times \sqrt{3} = \sqrt{2 \times 3} = \sqrt{6}}$

This multiplication of two square roots results in ${\sqrt{6}}$, which will be the constant term in the quadratic polynomial.

Step 3: Construct the Quadratic Polynomial

Using the general form of the quadratic polynomial with the sum and product of the roots, we have:

${x^2 - (\alpha + \beta)x + \alpha \beta}$

Substituting the values we calculated:

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

Thus, the required quadratic polynomial is ${x^2 - (\sqrt{2} + \sqrt{3})x + \sqrt{6}}$. This polynomial has ${\sqrt{2}}$ and ${\sqrt{3}}$ as its zeros, fulfilling the conditions of the problem.

Analyzing the Options

Given the options:

(1) ${x^2 + (\sqrt{2} + \sqrt{3})x + \sqrt{6}}$ (2) ${x^2 - (\sqrt{2} + \sqrt{3})x + \sqrt{6}}$ (3) ${x^2 + (\sqrt{2} + \sqrt{3})x - \sqrt{6}}$ (4) ${x^2 - (\sqrt{2} + \sqrt{3})x - \sqrt{6}}$

We can compare these options with the quadratic polynomial we derived:

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

Comparing with Option (1)

Option (1) is ${x^2 + (\sqrt{2} + \sqrt{3})x + \sqrt{6}}$. The coefficient of the ${x}$ term is positive, whereas in our derived polynomial, it is negative. Therefore, option (1) is incorrect.

Comparing with Option (2)

Option (2) is ${x^2 - (\sqrt{2} + \sqrt{3})x + \sqrt{6}}$. This matches exactly with our derived polynomial. Thus, option (2) is the correct answer.

Comparing with Options (3) and (4)

Options (3) and (4) have incorrect signs for the constant term. Option (3) is ${x^2 + (\sqrt{2} + \sqrt{3})x - \sqrt{6}}$, and option (4) is ${x^2 - (\sqrt{2} + \sqrt{3})x - \sqrt{6}}$. Both of these have a negative constant term, while our derived polynomial has a positive constant term ${\sqrt{6}}$. Therefore, options (3) and (4) are incorrect.

Conclusion

After analyzing all the options, it is clear that option (2) ${x^2 - (\sqrt{2} + \sqrt{3})x + \sqrt{6}}$ is the correct quadratic polynomial with roots ${\sqrt{2}}$ and ${\sqrt{3}}$.

Alternative Methods and Considerations

While the method of using the sum and product of roots is efficient, there are alternative ways to approach this problem and some considerations to keep in mind.

Method 1: Using the Factor Theorem

The Factor Theorem states that if ${\alpha}$ is a root of a polynomial ${P(x)}$, then ${(x - \alpha)}$ is a factor of ${P(x)}$. Given the roots ${\sqrt{2}}$ and ${\sqrt{3}}$, we can write the factors as ${(x - \sqrt{2})}$ and ${(x - \sqrt{3})}$. The quadratic polynomial can then be constructed by multiplying these factors:

${(x - \sqrt{2})(x - \sqrt{3})}$

Expanding this, we get:

${x^2 - x\sqrt{3} - x\sqrt{2} + \sqrt{2}\sqrt{3}}$

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

This method directly applies the Factor Theorem and arrives at the same quadratic polynomial as before.

Method 2: Verification by Substitution

Another way to verify the correctness of the polynomial is by substituting the roots back into the polynomial and checking if the result is zero. For the polynomial ${x^2 - (\sqrt{2} + \sqrt{3})x + \sqrt{6}}$, we substitute ${x = \sqrt{2}}$:

${(\sqrt{2})^2 - (\sqrt{2} + \sqrt{3})(\sqrt{2}) + \sqrt{6}}$

${2 - (2 + \sqrt{6}) + \sqrt{6}}$

${2 - 2 - \sqrt{6} + \sqrt{6} = 0}$

Similarly, substituting ${x = \sqrt{3}}$:

${(\sqrt{3})^2 - (\sqrt{2} + \sqrt{3})(\sqrt{3}) + \sqrt{6}}$

${3 - (\sqrt{6} + 3) + \sqrt{6}}$

${3 - \sqrt{6} - 3 + \sqrt{6} = 0}$

Both roots satisfy the polynomial, confirming its correctness.

Considerations

• Non-zero Constant Multiple: The quadratic polynomial is not unique; any non-zero constant multiple of the polynomial will also have the same roots. For example, ${2(x^2 - (\sqrt{2} + \sqrt{3})x + \sqrt{6}))}$ would also have ${\sqrt{2}}$ and ${\sqrt{3}}$ as roots.
• Complex Roots: If the roots were complex numbers, the same methods apply. The sum and product of complex roots can be used to construct the quadratic polynomial.

Common Pitfalls and Mistakes

When constructing quadratic polynomials from roots, several common mistakes can occur. Being aware of these pitfalls can help prevent errors.

Sign Errors

One of the most common mistakes is making errors with signs. In the general form ${x^2 - (\alpha + \beta)x + \alpha \beta}$, the sum of the roots has a negative sign in front of it. It's easy to forget this negative sign, especially when the roots themselves are negative.

Incorrect Multiplication

When calculating the product of roots, especially when dealing with square roots or complex numbers, it's important to be careful. Errors in multiplication can lead to an incorrect constant term in the polynomial.

Forgetting the Constant Multiple

Remember that the quadratic polynomial is not unique. Any non-zero constant multiple of the polynomial will have the same roots. While it's common to take the constant multiple as 1 for simplicity, it's important to recognize that other multiples are also valid.

Not Verifying the Result

It's always a good practice to verify the result by substituting the roots back into the polynomial. This can help catch errors in calculations or sign mistakes.

Misunderstanding the Factor Theorem

When using the Factor Theorem, ensure that the factors are correctly formed as ${(x - \alpha)}$ and ${(x - \beta)}$. Reversing the signs can lead to an incorrect polynomial.

By being mindful of these common pitfalls, students can improve their accuracy and confidence in solving problems involving quadratic polynomials.

Conclusion

In summary, finding a quadratic polynomial with given roots involves understanding the relationship between the roots and coefficients, and applying the appropriate formulas. For roots ${\sqrt{2}}$ and ${\sqrt{3}}$, the correct quadratic polynomial is ${x^2 - (\sqrt{2} + \sqrt{3})x + \sqrt{6}}$. This result is derived by calculating the sum and product of the roots and substituting them into the general form of a quadratic polynomial.

We also explored alternative methods, such as using the Factor Theorem and verifying the result by substitution. Additionally, we discussed common pitfalls and mistakes to avoid, such as sign errors and incorrect multiplication. By mastering these techniques and being aware of potential errors, one can confidently solve problems involving quadratic polynomials.

The understanding of quadratic polynomials is fundamental in algebra and has wide-ranging applications in various fields of mathematics and science. Whether it's solving equations, modeling physical phenomena, or tackling complex problems, the ability to work with quadratic polynomials is an essential skill.