Let's go through the details of each step that Roman followed to solve the quadratic equation by completing the square.
The original equation given is:
[tex]\[ 0 = -x^2 + 6x - 15 \][/tex]
### Step 1:
[tex]\[ -x^2 + 6x - 15 = 0 \][/tex]
To make the coefficient of [tex]\(x^2\)[/tex] positive, multiply the entire equation by [tex]\(-1\)[/tex]:
[tex]\[ x^2 - 6x + 15 = 0 \][/tex]
### Step 2:
Subtract 15 from both sides to isolate the terms involving [tex]\(x\)[/tex]:
[tex]\[ x^2 - 6x + 15 - 15 = 0 - 15 \][/tex]
[tex]\[ x^2 - 6x = -15 \][/tex]
### Step 3:
To complete the square, add [tex]\((\frac{b}{2})^2\)[/tex] to both sides where [tex]\(b\)[/tex] is the coefficient of [tex]\(x\)[/tex]. Here, [tex]\(b = -6\)[/tex], so [tex]\((\frac{-6}{2})^2 = 9\)[/tex]:
[tex]\[ x^2 - 6x + 9 = -15 + 9 \][/tex]
[tex]\[ x^2 - 6x + 9 = -6 \][/tex]
### Step 4:
Rewrite the left-hand side as the square of a binomial:
[tex]\[ (x - 3)^2 = -6 \][/tex]
### Step 5:
Take the square root of both sides:
[tex]\[ x - 3 = \pm \sqrt{-6} \][/tex]
### Step 6:
Solve for [tex]\(x\)[/tex] by adding 3 to both sides:
Instead, Roman subtracted 3 again, which was incorrect:
[tex]\[ x - 3 - 3 = -3 \pm \sqrt{-6} \][/tex]
The correct step should be:
[tex]\[ x - 3 = \pm i\sqrt{6} \][/tex]
The mistake Roman made was in Step 6. He should have added 3 to both sides of the equation to isolate [tex]\(x\)[/tex]:
[tex]\[ x = 3 \pm i\sqrt{6} \][/tex]
Therefore, the correct answer is:
Yes, in Step 6, he should have added 3 to both sides of the equation.
