Problem 102 - Precalculus Student Solutions Manual 5th

Question

Show that \(-3+4 i\) is a solution of the equation \(x^{2}+6 x+25=0\)

Step-by-Step Solution

Answer

\(-3 + 4i\) satisfies the equation \(x^2 + 6x + 25 = 0\).

Step 1: Substitute the complex number

Substitute \(-3 + 4i\) into the equation \(x^2 + 6x + 25 = 0\). We need to calculate \((-3 + 4i)^2 + 6(-3 + 4i) + 25\).

Step 2: Calculate \((x)^2\)

First calculate \((-3 + 4i)^2\). Use the formula \(a^2 + 2ab + b^2\) where \(a = -3\) and \(b = 4i\). \((-3 + 4i)^2 = (-3)^2 + 2(-3)(4i) + (4i)^2 = 9 - 24i - 16 = -7 - 24i\).

Step 3: Calculate \(6x\)

Next, calculate \(6(-3 + 4i)\). This gives \6(-3) + 6(4i) = -18 + 24i\.

Step 4: Add the components together

Now, add all the components together: \[ (-7 - 24i) + (-18 + 24i) + 25 \]. Combine the real parts and the imaginary parts separately: \[-7 - 18 + 25 = 0\] and \[-24i + 24i = 0\]. So, \(-7 - 24i + -18 + 24i + 25 = 0 + 0 = 0\).

Step 5: Conclusion

Since the left side equals the right side, \(-3 + 4i\) is a solution of the equation \(x^2 + 6x + 25 = 0\).

Key Concepts

Quadratic Equations

Quadratic equations are polynomials of degree 2. They have the general form \[ ax^2 + bx + c = 0 \] Here,

• \(x\) represents the variable or unknown.
• \(a\), \(b\), and \(c\) are constants with \(a eq 0\).

Solutions to quadratic equations can be found using:

• Factoring
• Completing the square
• Quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Quadratic equations can have real or complex solutions, depending on the value of the discriminant \( b^2 - 4ac \). If the discriminant is negative, the quadratic equation has complex solutions. Complex solutions come in conjugate pairs, such as \(a + bi\) and \(a - bi\). Complex solutions are essential for solving many higher-level math problems!

Complex Solutions

Complex solutions arise when the discriminant of a quadratic equation is negative (\( b^2 - 4ac < 0 \)). Complex numbers are written in the form \(a + bi\), where:

• \(a\) is the real part.
• \(bi\) is the imaginary part, where \(i\) is the imaginary unit satisfying \(i^2 = -1\).

To verify a complex solution of a quadratic equation, substitute the complex number back into the equation. Let's say we have \[ x^2 + 6x + 25 = 0 \] and a solution \( -3 + 4i \). First, plug \( -3 + 4i \) into the equation: \[ (-3 + 4i)^2 + 6(-3 + 4i) + 25 \] Simplify the expression step by step. It's like solving different parts of a puzzle:

• Calculate \( (-3 + 4i)^2 \)
• Calculate \( 6(-3 + 4i) \)
• Add all terms together

If the left side equals the right side, then the complex number is a valid solution.

Algebraic Substitution

Algebraic substitution is a method used to solve equations, including those with complex numbers. Here is how it works: Take each part step by step, substituting the specific values into the quadratic equation. For example, let’s use \( -3 + 4i \) and the equation \( x^2 + 6x + 25 = 0 \):

• Step 1: Substitute \( x \) with \( -3 + 4i \).
• Step 2: Calculate \( (-3 + 4i)^2 \). Use the formula \((a + b)^2 = a^2 + 2ab + b^2\): \[ (-3)^2 + 2(-3)(4i) + (4i)^2 = 9 - 24i - 16 = -7 - 24i \]
• Step 3: Calculate \( 6(-3 + 4i) \): \[ -18 + 24i \]
• Step 4: Add all components together: \[ (-7 - 24i) + (-18 + 24i) + 25 = 0 \]

When solving quadratic equations, it is crucial to handle each part carefully. Keeping track of real and imaginary parts helps to ensure the equality holds true. Substitution makes it easier to see why a solution works.