Problem 104 - Algebra A Combined Function

Question

Determine whether \(-1+i\) is a solution of \(x^{2}+2 x=-2\)

Step-by-Step Solution

Answer

Yes, \(-1+i\) is a solution of \(x^2+2x=-2\).

Step 1: Substitute the Potential Solution

Consider the quadratic equation \(x^{2}+2x=-2\). We are asked to check if \(x = -1+i\) is a solution. Substitute \(x = -1+i\) into the equation: \[ ((-1+i)^{2}+2(-1+i)) = -2. \]

Step 2: Calculate the Square of \(-1+i\)

Calculate \((-1+i)^{2}\): \[ (-1+i)^2 = (-1)^2 + 2(-1)(i) + i^2 = 1 - 2i - 1 = -2i. \]

Step 3: Calculate the Linear Term

Now, calculate \(2(-1+i)\): \[ 2(-1+i) = -2 + 2i. \]

Step 4: Combine Terms

Add the results from Steps 2 and 3: \[ -2i + (-2 + 2i) = -2i - 2 + 2i = -2. \]

Step 5: Verify the Result

Compare the result from Step 4 to the right side of the equation \(-2\). Since they are equal, the substitution satisfies the equation. Thus, \(-1+i\) is indeed a solution.

Key Concepts

Quadratic Equations

Quadratic equations are a key concept in algebra that involve equations expressed in the form \(ax^2 + bx + c = 0\). These equations characterize a parabola when graphed on a coordinate plane. The general solutions to quadratic equations can be found using the quadratic formula:

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

For an equation such as \(x^2 + 2x = -2\), you first bring it to the standard quadratic form by rearranging terms to get \(x^2 + 2x + 2 = 0\).
Here, \(a = 1\), \(b = 2\), and \(c = 2\). The solutions can be real or complex depending on the discriminant \(b^2 - 4ac\). If the discriminant is positive, we have two distinct real solutions; if it's zero, we have a repeated real solution; and if it's negative, the solutions are complex. This underscores the importance of understanding the nature of the discriminant in solving quadratic equations.

Polynomial Equations

Polynomial equations extend beyond quadratics and can involve terms with degrees higher than two. A polynomial equation is typically expressed as:

• \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0\)

Where \(n\) is a non-negative integer, and \(a_n\) through \(a_0\) are constants, with \(a_n eq 0\). The solutions to polynomial equations of higher degrees can be complex and often require numerical methods or factoring techniques.
When evaluating such equations as \(x^{2} + 2x = -2\), it becomes essential to consider not only real numbers but also complex numbers, which are often the solutions to these more advanced polynomial equations. Complex numbers are of the form \(a + bi\), where \(i\) is the imaginary unit. In this exercise, the potential solution \(-1 + i\) fits this description, highlighting the role of complex numbers in solving polynomial equations.

Solutions to Equations

Finding solutions to equations is crucial in mathematics, assisting in determining the points at which an equation's expression equals zero. Solutions can be numbers that satisfy the equation when substituted for the variable. For nonlinear equations, such as our quadratic example, the solutions might not always be immediately apparent or even real numbers.
Let's consider our specific example: a quadratic equation \(x^2 + 2x + 2 = 0\). The solution step involved substituting \(-1+i\) into the equation and simplifying systematically to verify if it results in the true statement

• If substitution holds, then it's indeed a solution.

This substitution process helps in confirming whether complex numbers like \(-1+i\) satisfy an equation. Such a procedure is a practical way to verify potential solutions, highlighting the need for understanding both computation and theory behind complex solutions.