Question

1. use substitution to verify each answer is a solution to the equation.

a. ( 6i ) is a solution to ( x^2 + 8 = -28 )
b. ( (3 - 4i) ) is a solution to ( x^2 - 6x + 25 = 0 )

Explanation:

Part a

Step1: Substitute \( x = 6i \) into left - hand side (LHS) of the equation \( x^{2}+8=-28 \)

We know that \( i^{2}=-1 \). So, if \( x = 6i \), then \( x^{2}=(6i)^{2}=6^{2}\times i^{2}=36\times(- 1)=-36 \)

Step2: Calculate LHS of the equation

LHS \(=x^{2}+8\). Substitute \( x^{2}=-36 \) into it, we get LHS \(=-36 + 8=-28 \)
The right - hand side (RHS) of the equation \( x^{2}+8=-28 \) is \(-28\). Since LHS = RHS, \( 6i \) is a solution of the equation \( x^{2}+8=-28 \)

Part b

Step1: Substitute \( x = 3 - 4i \) into the left - hand side (LHS) of the equation \( x^{2}-6x + 25 = 0 \)

First, calculate \( x^{2}=(3 - 4i)^{2}\). Using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\), where \( a = 3 \) and \( b = 4i \)
\(x^{2}=(3)^{2}-2\times3\times(4i)+(4i)^{2}=9-24i + 16i^{2}\)
Since \( i^{2}=-1 \), \( x^{2}=9-24i+16\times(-1)=9 - 24i-16=-7-24i \)
Then, calculate \( - 6x=-6\times(3 - 4i)=-18 + 24i \)

Step2: Calculate LHS of the equation

LHS \(=x^{2}-6x + 25\). Substitute \( x^{2}=-7-24i \), \( - 6x=-18 + 24i \) into it:
LHS \(=(-7-24i)+(-18 + 24i)+25\)
Combine like terms: \((-7-18 + 25)+(-24i + 24i)\)
\((-25 + 25)+0i=0\)
The right - hand side (RHS) of the equation \( x^{2}-6x + 25 = 0 \) is \( 0 \). Since LHS = RHS, \( 3 - 4i \) is a solution of the equation \( x^{2}-6x + 25 = 0 \)

Answer:

a. \( 6i \) is a solution of \( x^{2}+8=-28 \) (verified by substitution).
b. \( 3 - 4i \) is a solution of \( x^{2}-6x + 25 = 0 \) (verified by substitution).