Solve The Equation: ${(x+1)^2 + Y^2 + 6y = 5}$

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Introduction

Solving equations is a fundamental concept in mathematics, and it's essential to understand how to approach and solve various types of equations. In this article, we will focus on solving a quadratic equation in two variables, $(x+1)^2 + y^2 + 6y = 5$. This equation involves a squared term and a linear term, making it a quadratic equation in $y$. Our goal is to isolate the variable $y$ and find its possible values.

Understanding the Equation

The given equation is $(x+1)^2 + y^2 + 6y = 5$. To solve this equation, we need to understand the properties of quadratic equations and how to manipulate them to isolate the variable $y$. A quadratic equation in one variable has the general form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants. In our equation, we have a squared term $(x+1)^2$, which can be expanded to $x^2 + 2x + 1$. The linear term is $y^2 + 6y$, and the constant term is $5$.

Expanding the Squared Term

To simplify the equation, we can expand the squared term $(x+1)^2$ using the formula $(a+b)^2 = a^2 + 2ab + b^2$. In this case, $a = x$ and $b = 1$, so we have:

$(x+1)^2 = x^2 + 2x + 1$

Substituting this expansion into the original equation, we get:

$x^2 + 2x + 1 + y^2 + 6y = 5$

Simplifying the Equation

Now, we can simplify the equation by combining like terms. We have $x^2 + 2x + 1$ on the left-hand side, and $y^2 + 6y$ on the left-hand side. We can rewrite the equation as:

$x^2 + 2x + 1 + y^2 + 6y = 5$

$x^2 + 2x + 1 + y^2 + 6y - 5 = 0$

$x^2 + 2x + 1 + y^2 + 6y - 5 = 0$

Completing the Square

To complete the square, we need to add and subtract a constant term to make the left-hand side a perfect square trinomial. In this case, we can add and subtract $9$ to make the left-hand side a perfect square trinomial:

$x^2 + 2x + 1 + y^2 + 6y - 5 + 9 - 9 = 0$

$x^2 + 2x + 1 + (y^2 + 6y + 9) - 9 - 5 = 0$

$(x+1)^2 + (y+3)^2 - 14 = 0$

Isolating the Variable $y$

Now, we can isolate the variable $y$ by adding $14$ to both sides of the equation:

$(x+1)^2 + (y+3)^2 = 14$

Subtracting $(x+1)^2$ from both sides, we get:

$(y+3)^2 = 14 - (x+1)^2$

Taking the square root of both sides, we get:

$y+3 = \pm \sqrt{14 - (x+1)^2}$

Subtracting $3$ from both sides, we get:

$y = -3 \pm \sqrt{14 - (x+1)^2}$

Conclusion

In this article, we solved the equation $(x+1)^2 + y^2 + 6y = 5$ by expanding the squared term, simplifying the equation, completing the square, and isolating the variable $y$. We found that the solution to the equation is $y = -3 \pm \sqrt{14 - (x+1)^2}$. This equation represents a pair of parabolas that intersect at the point $(0, -2)$.

Final Answer

The final answer to the equation $(x+1)^2 + y^2 + 6y = 5$ is:

$y = -3 \pm \sqrt{14 - (x+1)^2}$

This equation represents a pair of parabolas that intersect at the point $(0, -2)$.

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Introduction

In our previous article, we solved the equation $(x+1)^2 + y^2 + 6y = 5$ by expanding the squared term, simplifying the equation, completing the square, and isolating the variable $y$. We found that the solution to the equation is $y = -3 \pm \sqrt{14 - (x+1)^2}$. In this article, we will answer some common questions related to the solution of this equation.

Q&A

Q: What is the final answer to the equation $(x+1)^2 + y^2 + 6y = 5$?

A: The final answer to the equation $(x+1)^2 + y^2 + 6y = 5$ is $y = -3 \pm \sqrt{14 - (x+1)^2}$.

Q: What is the significance of the $\pm$ symbol in the solution?

A: The $\pm$ symbol in the solution indicates that there are two possible values for $y$ for each value of $x$. This is because the square root of a number can be either positive or negative.

Q: What is the shape of the graph of the equation $(x+1)^2 + y^2 + 6y = 5$?

A: The graph of the equation $(x+1)^2 + y^2 + 6y = 5$ is a pair of parabolas that intersect at the point $(0, -2)$.

Q: How do I graph the equation $(x+1)^2 + y^2 + 6y = 5$?

A: To graph the equation $(x+1)^2 + y^2 + 6y = 5$, you can use a graphing calculator or a computer algebra system. You can also use a graphing software such as Desmos or GeoGebra.

Q: What is the domain and range of the equation $(x+1)^2 + y^2 + 6y = 5$?

A: The domain of the equation $(x+1)^2 + y^2 + 6y = 5$ is all real numbers, and the range is all real numbers.

Q: How do I find the intersection points of the two parabolas?

A: To find the intersection points of the two parabolas, you can set the two equations equal to each other and solve for $x$. Then, you can substitute the value of $x$ into one of the equations to find the corresponding value of $y$.

Q: What is the significance of the point $(0, -2)$ in the graph of the equation $(x+1)^2 + y^2 + 6y = 5$?

A: The point $(0, -2)$ is the intersection point of the two parabolas. It is also the vertex of the parabola $y = -3 \pm \sqrt{14 - (x+1)^2}$.

Conclusion

In this article, we answered some common questions related to the solution of the equation $(x+1)^2 + y^2 + 6y = 5$. We found that the solution to the equation is $y = -3 \pm \sqrt{14 - (x+1)^2}$, and we discussed the significance of the $\pm$ symbol, the shape of the graph, and the domain and range of the equation. We also provided some tips on how to graph the equation and find the intersection points of the two parabolas.

Final Answer

The final answer to the equation $(x+1)^2 + y^2 + 6y = 5$ is:

$y = -3 \pm \sqrt{14 - (x+1)^2}$

This equation represents a pair of parabolas that intersect at the point $(0, -2)$.