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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][tex]$y$[/tex][/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]$(b \div 2)^2[/tex] for the [tex]$x$[/tex] and [tex]$y$[/tex] terms.
[tex]
(12 \div 2)^2=36 \quad \text{and} \quad (2 \div 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=38
[/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's complete the final step of converting the given equation to standard form by completing the square.

4. Write each trinomial as a binomial squared, and simplify the right side.

We start with the equation from step 3:
[tex]\[ x^2 + 12x + 36 + y^2 + 2y + 1 = 1 + 36 + 1 \][/tex]

Now, express each trinomial as a binomial squared:
[tex]\[ x^2 + 12x + 36 = (x + 6)^2 \][/tex]
[tex]\[ y^2 + 2y + 1 = (y + 1)^2 \][/tex]

So, the left side of the equation becomes:
[tex]\[ (x + 6)^2 + (y + 1)^2 \][/tex]

The right side simplifies to:
[tex]\[ 1 + 36 + 1 = 38 \][/tex]

Therefore, the equation in standard form is:
[tex]\[ (x + 6)^2 + (y + 1)^2 = 38 \][/tex]

So, [tex]\((x + \boxed{6})^2 + (y + \boxed{1})^2 = 38\)[/tex].
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