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There are four steps for converting the equation [tex]\(x^2+y^2+12x+2y-1=0\)[/tex] into standard form by completing the square. Complete the last step.

1. Group the [tex]\(x\)[/tex] terms together and the [tex]\(y\)[/tex] terms together, and move the constant term to the other side of the equation.
[tex]\[x^2 + 12x + y^2 + 2y = 1\][/tex]

2. Determine [tex]\((b/2)^2\)[/tex] for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
[tex]\[(12/2)^2 = 36 \quad \text{and} \quad (2/2)^2 = 1\][/tex]

3. Add the values to both sides of the equation.
[tex]\[x^2 + 12x + 36 + y^2 + 2y + 1 = 1 + 36 + 1\][/tex]

4. Write each trinomial as a binomial squared, and simplify the right side.
[tex]\[(x + \square)^2 + (y + \square)^2 = \square\][/tex]

Complete the last step.

Sagot :

GinnyAnsweridn
We will now complete the last step step-by-step to convert the given equation into its standard form by completing the square.

Given that we have completed steps 1 through 3, we have:
[tex]\[ x^2 + 12x + 36 + y^2 + 2y + 1 = 1 + 36 + 1 \][/tex]

### Step 4: Write each trinomial as a binomial squared and simplify the right side

First, let's factor the trinomials on the left side:
[tex]\[ x^2 + 12x + 36 = (x + 6)^2 \][/tex]
[tex]\[ y^2 + 2y + 1 = (y + 1)^2 \][/tex]

So, the equation:
[tex]\[ x^2 + 12x + 36 + y^2 + 2y + 1 = 38 \][/tex]
becomes:
[tex]\[ (x + 6)^2 + (y + 1)^2 = 38 \][/tex]

Thus, the completed squares and the standard form of the given equation are:
[tex]\[ (x + 6)^2 + (y + 1)^2 = 38 \][/tex]

So, completing the square, we have:
[tex]\[ (x + 6), (y + 1), 38 \][/tex]
Therefore, the final equation in its standard form is:
[tex]\[ (x + 6)^2 + (y + 1)^2 = 38 \][/tex]
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