Problem 101
Question
Show that \(-2+i\) is a solution of the equation \(x^{2}+4 x+5=0\)
Step-by-Step Solution
Verified Answer
Yes, \( -2 + i \) is a solution since it satisfies \( x^2 + 4x + 5 = 0 \).
1Step 1 - Identify the Polynomial and the Given Solution
The polynomial given in the exercise is: \[ f(x) = x^2 + 4x + 5 \] The provided solution to verify is: \[ x = -2 + i \]
2Step 2 - Substitute the Solution into the Polynomial
To verify if \( -2 + i \) is a solution, substitute \( x = -2 + i \) into the polynomial: \[ f(-2+i) = (-2 + i)^2 + 4(-2 + i) + 5 \]
3Step 3 - Compute the Square of the Complex Number
Calculate \( (-2 + i)^2 \) using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \): \[ (-2 + i)^2 = (-2)^2 + 2(-2)(i) + (i)^2 \] This simplifies to: \[ 4 - 4i + i^2 \] Since \( i^2 = -1 \), the expression becomes: \[ 4 - 4i - 1 = 3 - 4i \]
4Step 4 - Compute the Linear Term
Calculate the term \( 4(-2 + i) \): \[ 4(-2 + i) = 4(-2) + 4(i) \] This simplifies to: \[ -8 + 4i \]
5Step 5 - Combine All the Terms
Combine all terms calculated into the polynomial expression: \[ f(-2+i) = (3 - 4i) + (-8 + 4i) + 5 \] Simplify by combining like terms: \[ 3 - 4i - 8 + 4i + 5 \] The imaginary parts cancel out, and the expression becomes: \[ 0 \]
Key Concepts
Polynomial EquationsComplex NumbersSubstitution MethodSimplification
Polynomial Equations
A polynomial equation involves variables raised to whole number exponents and can have multiple terms.
The general form of a polynomial equation is: \[ a_n x^n + a_{n-1} x^{n-1} + \text{...} + a_1 x + a_0 = 0 \] where \(a_i\) are coefficients and \(n\) is a non-negative integer.
In our exercise, the polynomial is \(x^2 + 4x + 5 = 0\).
Understanding polynomial equations requires recognizing key parts:
The general form of a polynomial equation is: \[ a_n x^n + a_{n-1} x^{n-1} + \text{...} + a_1 x + a_0 = 0 \] where \(a_i\) are coefficients and \(n\) is a non-negative integer.
In our exercise, the polynomial is \(x^2 + 4x + 5 = 0\).
Understanding polynomial equations requires recognizing key parts:
- Terms Polynomial terms consist of variables and coefficients.
- Degree The degree is the highest exponent. Here, \(x^2\) means the polynomial is of degree 2.
- Solutions These are values for which the polynomial equals zero.
Our task is to verify if \(-2+i\) is a solution.
Complex Numbers
Complex numbers include real and imaginary parts.
A complex number is expressed as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part.
Imaginary numbers involve the unit \(i\), defined as \(i = \sqrt{-1}\), and \(i^2 = -1\).
In the exercise, our solution \(-2+i\) has real part \(-2\) and imaginary part \(i\).
Complex numbers are crucial because they can be solutions to polynomial equations, especially those of higher degree.
Substitution of complex roots verifies solutions by ensuring the polynomial evaluates to zero.
A complex number is expressed as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part.
Imaginary numbers involve the unit \(i\), defined as \(i = \sqrt{-1}\), and \(i^2 = -1\).
In the exercise, our solution \(-2+i\) has real part \(-2\) and imaginary part \(i\).
Complex numbers are crucial because they can be solutions to polynomial equations, especially those of higher degree.
Substitution of complex roots verifies solutions by ensuring the polynomial evaluates to zero.
Substitution Method
The substitution method involves replacing the variable in an equation with a given value.
For our polynomial \(x^2 + 4x + 5\), we substitute \(x\) with \(-2+i\).
Steps for substitution:
In this exercise, we see:
\(f(-2+i) = (-2 + i)^2 + 4(-2 + i) + 5\)
For our polynomial \(x^2 + 4x + 5\), we substitute \(x\) with \(-2+i\).
Steps for substitution:
- Step 1 Identify the value to substitute (\(-2+i\)).
- Step 2 Replace all instances of \(x\) in the polynomial with \(-2+i\).
- Step 3 Calculate by handling complex number arithmetic.
In this exercise, we see:
\(f(-2+i) = (-2 + i)^2 + 4(-2 + i) + 5\)
Simplification
Simplification is about reducing expressions to their simplest forms.
In our polynomial, after substitution, we simplify \( (-2+i)^2 \) and \(4(-2+i)\).
Calculations: \
In our polynomial, after substitution, we simplify \( (-2+i)^2 \) and \(4(-2+i)\).
Calculations: \
- \((-2+i)^2 = (-2)^2 + 2(-2)(i) + (i)^2 \text=> 4 - 4i - 1 = 3 - 4i\)
- \(4(-2+i) = -8 + 4i\)
- \(f(-2+i) = (3-4i) + (-8+4i) + 5\)
- \(3-4i-8+4i+5\)
- The imaginary parts cancel out, simplifying to \(0\).
Other exercises in this chapter
Problem 98
Find the values of \(a, b,\) and \(c\) for which the quadratic equation $$ a x^{2}+b x+c=0 $$ has the given numbers as solutions. (Hint: Use the zero-factor pro
View solution Problem 100
Solve each equation for the indicated variable. Assume all denominators are nonzero. $$\frac{1}{R}=\frac{1}{r_{1}}+\frac{1}{r_{2}}, \quad \text { for } R$$
View solution Problem 101
Solve each equation for the indicated variable. Assume all denominators are nonzero. $$\frac{E}{e}=\frac{R+r}{r}, \quad \text { for } e$$
View solution Problem 102
Show that \(-3+4 i\) is a solution of the equation \(x^{2}+6 x+25=0\)
View solution