What Are The Solutions Of The Equation \[$(x+2)^2-2(x+2)-15=0\$\]?Use \[$u\$\] Substitution To Solve.A. \[$x=-7\$\] And \[$x=1\$\] B. \[$x=-5\$\] And \[$x=3\$\] C. \[$x=-3\$\] And

$$$$

Introduction

Solving quadratic equations can be a challenging task, especially when they are not in the standard form. In this article, we will explore the solution to the equation {(x+2)^2-2(x+2)-15=0$}$ using the substitution method. This method involves replacing a variable in the equation with a new variable, which simplifies the equation and makes it easier to solve.

The Equation

The given equation is {(x+2)^2-2(x+2)-15=0$}$. This equation is not in the standard form, which is {ax^2+bx+c=0$}$. To solve this equation, we need to simplify it and rewrite it in the standard form.

Using the Substitution Method

To simplify the equation, we can use the substitution method. Let's substitute {u=x+2$}$ into the equation. This means that we will replace {x+2$}$ with {u$}$ in the equation.

Substituting {u=x+2$}$ into the Equation

When we substitute {u=x+2$}$ into the equation, we get:

{u^2-2u-15=0$}$

This equation is now in the standard form, which is {ax^2+bx+c=0$}$. We can see that the coefficient of {u^2$}$ is 1, the coefficient of {u$}$ is -2, and the constant term is -15.

Solving the Equation

To solve the equation {u^2-2u-15=0$}$, we can use the quadratic formula. The quadratic formula is:

{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$}$

In this case, {a=1$}$, {b=-2$}$, and {c=-15$}$. Plugging these values into the quadratic formula, we get:

{u=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-15)}}{2(1)}$}$

Simplifying the expression, we get:

{u=\frac{2\pm\sqrt{4+60}}{2}$}$

{u=\frac{2\pm\sqrt{64}}{2}$}$

{u=\frac{2\pm8}{2}$}$

This gives us two possible values for {u$}$:

{u=\frac{2+8}{2}=5$}$

{u=\frac{2-8}{2}=-3$}$

Back Substitution

Now that we have found the values of {u$}$, we can substitute them back into the equation {u=x+2$}$ to find the values of {x$}$.

For {u=5$}$, we have:

${5=x+2\$}

Subtracting 2 from both sides, we get:

${3=x\$}

So, the first solution is {x=3$}$.

For {u=-3$}$, we have:

{-3=x+2$}$

Subtracting 2 from both sides, we get:

{-5=x$}$

So, the second solution is {x=-5$}$.

Conclusion

In this article, we used the substitution method to solve the equation {(x+2)^2-2(x+2)-15=0$}$. We substituted {u=x+2$}$ into the equation and simplified it to get {u^2-2u-15=0$}$. We then used the quadratic formula to solve the equation and found two possible values for {u$}$. Finally, we substituted these values back into the equation {u=x+2$}$ to find the values of {x$}$. The solutions to the equation are {x=-5$}$ and {x=3$}$.

Final Answer

The final answer is: B. {x=-5$}$ and {x=3$}$

$$$$

Introduction

Solving quadratic equations can be a challenging task, especially when they are not in the standard form. In this article, we will explore the solution to the equation {(x+2)^2-2(x+2)-15=0$}$ using the substitution method. This method involves replacing a variable in the equation with a new variable, which simplifies the equation and makes it easier to solve.

The Equation

The given equation is {(x+2)^2-2(x+2)-15=0$}$. This equation is not in the standard form, which is {ax^2+bx+c=0$}$. To solve this equation, we need to simplify it and rewrite it in the standard form.

Using the Substitution Method

To simplify the equation, we can use the substitution method. Let's substitute {u=x+2$}$ into the equation. This means that we will replace {x+2$}$ with {u$}$ in the equation.

Substituting {u=x+2$}$ into the Equation

When we substitute {u=x+2$}$ into the equation, we get:

{u^2-2u-15=0$}$

This equation is now in the standard form, which is {ax^2+bx+c=0$}$. We can see that the coefficient of {u^2$}$ is 1, the coefficient of {u$}$ is -2, and the constant term is -15.

Solving the Equation

To solve the equation {u^2-2u-15=0$}$, we can use the quadratic formula. The quadratic formula is:

{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$}$

In this case, {a=1$}$, {b=-2$}$, and {c=-15$}$. Plugging these values into the quadratic formula, we get:

{u=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-15)}}{2(1)}$}$

Simplifying the expression, we get:

{u=\frac{2\pm\sqrt{4+60}}{2}$}$

{u=\frac{2\pm\sqrt{64}}{2}$}$

{u=\frac{2\pm8}{2}$}$

This gives us two possible values for {u$}$:

{u=\frac{2+8}{2}=5$}$

{u=\frac{2-8}{2}=-3$}$

Back Substitution

Now that we have found the values of {u$}$, we can substitute them back into the equation {u=x+2$}$ to find the values of {x$}$.

For {u=5$}$, we have:

${5=x+2\$}

Subtracting 2 from both sides, we get:

${3=x\$}

So, the first solution is {x=3$}$.

For {u=-3$}$, we have:

{-3=x+2$}$

Subtracting 2 from both sides, we get:

{-5=x$}$

So, the second solution is {x=-5$}$.

Conclusion

In this article, we used the substitution method to solve the equation {(x+2)^2-2(x+2)-15=0$}$. We substituted {u=x+2$}$ into the equation and simplified it to get {u^2-2u-15=0$}$. We then used the quadratic formula to solve the equation and found two possible values for {u$}$. Finally, we substituted these values back into the equation {u=x+2$}$ to find the values of {x$}$. The solutions to the equation are {x=-5$}$ and {x=3$}$.

Final Answer

The final answer is: B. {x=-5$}$ and {x=3$}$

Q&A

Q: What is the substitution method in solving quadratic equations?

A: The substitution method involves replacing a variable in the equation with a new variable, which simplifies the equation and makes it easier to solve.

Q: Why do we need to simplify the equation before solving it?

A: We need to simplify the equation before solving it because it makes the equation easier to solve and reduces the chances of making mistakes.

Q: What is the quadratic formula?

A: The quadratic formula is a formula used to solve quadratic equations. It is given by:

{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$}$

Q: How do we use the quadratic formula to solve the equation?

A: To use the quadratic formula, we need to plug in the values of {a$}$, {b$}$, and {c$}$ into the formula and simplify the expression.

Q: What are the two possible values for {u$}$?

A: The two possible values for {u$}$ are {u=5$}$ and {u=-3$}$.

Q: How do we find the values of {x$}$ from the values of {u$}$?

A: We find the values of {x$}$ by substituting the values of {u$}$ back into the equation {u=x+2$}$ and solving for {x$}$.

Q: What are the solutions to the equation?

A: The solutions to the equation are {x=-5$}$ and {x=3$}$.

Q: Why is the substitution method useful in solving quadratic equations?

A: The substitution method is useful in solving quadratic equations because it simplifies the equation and makes it easier to solve.

Q: Can we use the substitution method to solve any quadratic equation?

A: No, we cannot use the substitution method to solve any quadratic equation. The substitution method is only useful when the equation can be simplified by substituting a new variable.

Q: What are some other methods for solving quadratic equations?

A: Some other methods for solving quadratic equations include factoring, completing the square, and using the quadratic formula.

Q: How do we choose which method to use when solving a quadratic equation?

A: We choose which method to use based on the complexity of the equation and the ease of use of each method.

Q: Can we use the substitution method to solve equations with complex coefficients?

A: No, we cannot use the substitution method to solve equations with complex coefficients. The substitution method is only useful for solving equations with real coefficients.

Q: What are some common mistakes to avoid when using the substitution method?

A: Some common mistakes to avoid when using the substitution method include substituting the wrong value for {u$}$, not simplifying the equation enough, and not checking for extraneous solutions.

Q: How do we check for extraneous solutions when using the substitution method?

A: We check for extraneous solutions by substituting the value of {x$}$ back into the original equation and checking if it is true.

Q: What are some real-world applications of the substitution method?

A: Some real-world applications of the substitution method include solving problems in physics, engineering, and economics.

Q: Can we use the substitution method to solve systems of equations?

A: Yes, we can use the substitution method to solve systems of equations. We can substitute one equation into the other equation and solve for the variables.

Q: How do we choose which variable to substitute when solving a system of equations?

A: We choose which variable to substitute based on the ease of use of each method and the complexity of the equations.

Q: What are some other methods for solving systems of equations?

A: Some other methods for solving systems of equations include substitution, elimination, and graphing.

Q: How do we choose which method to use when solving a system of equations?

A: We choose which method to use based on the complexity of the equations and the ease of use of each method.

Q: Can we use the substitution method to solve equations with multiple variables?

A: Yes, we can use the substitution method to solve equations with multiple variables. We can substitute one variable into the other equation and solve for the variables.

Q: How do we choose which variable to substitute when solving an equation with multiple variables?

A: We choose which variable to substitute based on the ease of use of each method and the complexity of the equations.

Q: What are some common mistakes to avoid when using the substitution method to solve equations with multiple variables?

A: Some common mistakes to avoid when using the substitution method to solve equations with multiple variables include substituting the wrong value for the variable, not simplifying the equation enough, and not checking for extraneous solutions.

Q: How do we check for extraneous solutions when using the substitution method to solve equations with multiple variables?

A: We check for extraneous solutions by substituting the value of the variable back into the original equation and checking if it is true.

Q: What are some real-world applications of the substitution method to solve equations with multiple variables?

A: Some real-world applications of the substitution method to solve equations with multiple variables include solving problems in physics, engineering