Question

Taking \(\mathrm{U}=[1,5], \mathrm{A}=\left\\{\mathrm{x} / \mathrm{x} \in \mathrm{N}, \mathrm{x}^{2}-6 \mathrm{x}+5=0\right\\} \mathrm{A}^{\prime}=\ldots\) (a) \(\\{1,5\\}\) (b) \((1,5)\) (c) \([1,5]\) (d) \([-1,-5]\)

Short answer

The short answer is \(A' = (1, 5)\).

Step-by-step solution

Solve the given equation for x

We need to find the values of x in the given equation: \(x^2 - 6x + 5 = 0\) This is a quadratic equation, which can be solved using the following formula to find the roots: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Here, a = 1, b = -6, and c = 5. We substitute these values into the formula and solve for x: \(x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)}\) \(x = \frac{6 \pm \sqrt{36-20}}{2}\) \(x = \frac{6 \pm \sqrt{16}}{2}\) Since \(x \in N\), we have: \(x = \frac{6 \pm 4}{2}\) This gives us two possible values for x: \\ 1) x = \(\frac{6+4}{2} = \frac{10}{2} = 5\) \\ 2) x = \(\frac{6-4}{2} = \frac{2}{2} = 1\) So our set is A = {1, 5}.

Find the complement of set A in set U

Now we need to find the complement of the set A in set U. Since set U = [1, 5], the set A' contains all the elements in U that are not in A. Since A = {1, 5}, the complement of A in U (A') will include everything within the interval except the elements 1 and 5 themselves. Thus: \(A' = (1, 5)\) Our final answer is A' = (1, 5), which matches with option (b).

Key concepts

Complement of a Set

Understanding the complement of a set is crucial in the realm of set theory and mathematics as a whole. The complement of a set, denoted as A', is essentially all of the elements that are not in set A but are within a given universal set U. It's important to visualize this with an example. Take the universal set U as all numbers from 1 to 5 inclusive, so U = [1, 5]. If we have a set A that contains specific elements from this universal set, say A = {2, 4}, then the complement of A (\( A' \) would be all the elements from U that are not in A. In this case, A' would be {1, 3, 5}.

To determine the complement of a set accurately, it's essential to start by defining the universal set U clearly, and then identify the members of set A. Whatever is left in U after removing A represents the complement of A. This concept can be applied in various areas such as probability, logic, and algebra to find solutions to complex problems.

Quadratic Formula

The quadratic formula — a reliable method for solving quadratic equations — is at the heart of algebra. An equation of the form \( ax^2 + bx + c = 0 \), where a, b, and c are constants, is a quadratic equation. The quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

unlocks the values of x that satisfy this equation, giving us the roots. The expression under the square root, \( b^2 - 4ac \), is known as the discriminant. It determines the nature of the roots — whether they're real or complex, and if real, whether they are distinct or identical. Knowing how to apply this formula is essential for anyone venturing into algebraic equations or higher mathematics. Not only does it provide answers, but it also deepens the understanding of functions and graphs related to quadratic equations.

Roots of a Polynomial

Roots of a polynomial are the solutions to the polynomial equation. For quadratic polynomials, which take the general form \( ax^2 + bx + c = 0 \), the roots are the values of x that make the equation true. These roots can be real or complex numbers and are found using various methods, including factoring, completing the square, or applying the quadratic formula. The roots of a polynomial are of profound significance in mathematics since they provide insights into the behavior of the graph of the polynomial function; for example, roots indicate where the graph intersects the x-axis.

It's fascinating to observe that the sum and product of the roots can be directly calculated from the coefficients of the quadratic equation, revealing intrinsic symmetries and properties of polynomials. For instance, in a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is \( -b/a \) and the product is \( c/a \). Understanding the roots is not just a procedural skill but also a foundation for further exploration in mathematics, including calculus and beyond.