Problem 89 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find two quadratic equations having the given solutions. (There are many correct answers.) $$2+i, 2-i$$

Step-by-Step Solution

Answer

Two quadratic equations having the given solutions are \( x^2 - 4x + 5 = 0 \) or \( x^2 - 4x + 5 = 0 \). Both equations give the roots \( 2+i \) and \( 2-i \).

Step 1: Derive the quadratic equation from the roots

The quadratic equation with roots \( r_1 \) and \( r_2 \) can be calculated as \( (x - r_1)(x - r_2) = 0 \). Given the two roots \( 2+i \) and \( 2-i \), the quadratic equation can be derived: \( (x - (2+i))(x - (2-i)) = 0 \).

Step 2: Expand the equation

Expanding the factors gives \( x^2 - 2x - ix - 2x + 4 + 2i - ix + 2i + 1 = 0 \). Simplifying this gives \( x^2 - 4x + 5 = 0 \).

Step 3: Check the roots

It can be verified that this quadratic equation has the given roots \( 2+i \) and \( 2-i \) by plugging these values into the equation. For example, for \( x = 2+i \), \( (2+i)^2 - 4(2+i) + 5 = 0 \), which holds true. Similarly, for \( x = 2-i \), \( (2-i)^2 - 4(2-i) + 5 = 0 \), which also holds true.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two, often written in the standard form: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). The shape of their graph is a parabola, which can open upwards or downwards depending on the sign of \( a \). Quadratic equations can have real or complex roots. **Complex roots** occur when the discriminant, \( b^2 - 4ac \), is negative. When solving quadratic equations, one common method is using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Because complex numbers can include imaginary parts, the solution often involves combining the real and imaginary components, like the given roots \( 2+i \) and \( 2-i \). Moreover, quadratics can be expressed in vertex form or by factoring, especially when the equation can be constructed from known roots.

Conjugate Roots Theorem

The Conjugate Roots Theorem is a pivotal concept in algebra, especially when dealing with polynomials and complex numbers. This theorem states that for any polynomial equation with real coefficients, if a complex number \( a+bi \) is a root, then its conjugate \( a-bi \) is also a root. This ensures that the non-real roots occur in pairs.By understanding this theorem, when given a root such as \( 2+i \), we can immediately infer that \( 2-i \) is also a root. This understanding is crucial when determining the quadratic equation from its roots. Knowing both roots allows us to form the polynomial by calculating **\((x - (a+bi))(x - (a-bi))\)**.

• This simplifies the process of multiplying binomials involving complex conjugates, as the imaginary parts \( (bi) \) cancel each other out.
• Thus, we get a real quadratic equation \((x - (2+i))(x - (2-i))\).

Applying this theorem reduces potential errors and makes the solution more straightforward when handling complex roots.

Polynomial Expansion

Polynomial expansion is a technique used to express products of polynomials, such as binomials or trinomials, as a simplified polynomial. This method is valuable when deriving equations like the quadratic equation from given roots.To expand the expression \((x - (2+i))(x - (2-i))\), the **FOIL method** (First, Outer, Inner, Last) is often used:

• First: Multiply the first terms, \( x \times x = x^2 \)
• Outer: Multiply the outer terms, \( x \times -(2-i) = -2x + ix \)
• Inner: Multiply the inner terms, \(-(2+i) \times x = -2x - ix \)
• Last: Multiply the last terms, \(-(2+i) \times -(2-i) = 4 + 1 \)

Then, the expression is combined and simplified: \( x^2 - 4x + 5 \).Knowing how to accurately expand polynomial expressions is essential for verifying the structure of quadratic equations. This step ensures the solutions align with the original complex roots provided.