What Is The Solution To The Equation X 2 + 2 X − 25 = X + 5 \sqrt{x^2+2x-25}=\sqrt{x+5} X 2 + 2 X − 25 ​ = X + 5 ​ ? Assume The Range Is All Real Numbers.A. X = − 6 X=-6 X = − 6 B. X = 5 X=5 X = 5 C. X = 6 , X = − 5 X=6, X=-5 X = 6 , X = − 5 D. X = − 6 , X = 5 X=-6, X=5 X = − 6 , X = 5

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Introduction

Solving equations involving square roots can be a challenging task, especially when the equation involves a quadratic expression inside the square root. In this article, we will explore the solution to the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$, assuming the range is all real numbers.

Understanding the Equation

The given equation is $\sqrt{x^2+2x-25}=\sqrt{x+5}$. To solve this equation, we need to first simplify the expressions inside the square roots. We can start by factoring the quadratic expression $x^2+2x-25$.

Factoring the Quadratic Expression

The quadratic expression $x^2+2x-25$ can be factored as $(x+5)(x-5)$. Therefore, the equation becomes $\sqrt{(x+5)(x-5)}=\sqrt{x+5}$.

Simplifying the Equation

Since the square root of a product is equal to the product of the square roots, we can simplify the equation as $\sqrt{x+5}\sqrt{x-5}=\sqrt{x+5}$. We can cancel out the common factor $\sqrt{x+5}$ from both sides of the equation, assuming that $x+5 \neq 0$.

Canceling Out the Common Factor

After canceling out the common factor $\sqrt{x+5}$, the equation becomes $\sqrt{x-5}=1$. To solve for $x$, we need to square both sides of the equation.

Squaring Both Sides

Squaring both sides of the equation $\sqrt{x-5}=1$ gives us $x-5=1$. We can add 5 to both sides of the equation to solve for $x$.

Solving for $x$

Adding 5 to both sides of the equation $x-5=1$ gives us $x=6$. Therefore, one possible solution to the equation is $x=6$.

Checking the Solution

To check if $x=6$ is a valid solution, we need to substitute it back into the original equation. Substituting $x=6$ into the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$ gives us $\sqrt{6^2+2(6)-25}=\sqrt{6+5}$, which simplifies to $\sqrt{36+12-25}=\sqrt{11}$, and further simplifies to $\sqrt{23}=\sqrt{11}$. Since $\sqrt{23} \neq \sqrt{11}$, $x=6$ is not a valid solution.

Revisiting the Equation

Since $x=6$ is not a valid solution, we need to revisit the equation and find the correct solution. We can start by re-examining the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$.

Re-examining the Equation

The equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$ can be rewritten as $\sqrt{(x+5)(x-5)}=\sqrt{x+5}$. We can simplify this equation by canceling out the common factor $\sqrt{x+5}$ from both sides of the equation.

Canceling Out the Common Factor

After canceling out the common factor $\sqrt{x+5}$, the equation becomes $\sqrt{x-5}=1$. To solve for $x$, we need to square both sides of the equation.

Squaring Both Sides

Squaring both sides of the equation $\sqrt{x-5}=1$ gives us $x-5=1$. We can add 5 to both sides of the equation to solve for $x$.

Solving for $x$

Adding 5 to both sides of the equation $x-5=1$ gives us $x=6$. However, we already know that $x=6$ is not a valid solution. Therefore, we need to find another solution.

Finding Another Solution

To find another solution, we can re-examine the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$. We can start by squaring both sides of the equation.

Squaring Both Sides

Squaring both sides of the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$ gives us $x^2+2x-25=x+5$. We can simplify this equation by combining like terms.

Simplifying the Equation

Combining like terms in the equation $x^2+2x-25=x+5$ gives us $x^2+x-30=0$. We can factor this quadratic equation as $(x+6)(x-5)=0$.

Factoring the Quadratic Equation

The quadratic equation $x^2+x-30=0$ can be factored as $(x+6)(x-5)=0$. Therefore, the solutions to the equation are $x=-6$ and $x=5$.

Checking the Solutions

To check if $x=-6$ and $x=5$ are valid solutions, we need to substitute them back into the original equation. Substituting $x=-6$ into the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$ gives us $\sqrt{(-6)^2+2(-6)-25}=\sqrt{-6+5}$, which simplifies to $\sqrt{36-12-25}=\sqrt{-1}$, and further simplifies to $\sqrt{(-1)}=\sqrt{-1}$. Since $\sqrt{(-1)}=\sqrt{-1}$, $x=-6$ is a valid solution. Substituting $x=5$ into the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$ gives us $\sqrt{5^2+2(5)-25}=\sqrt{5+5}$, which simplifies to $\sqrt{25+10-25}=\sqrt{10}$, and further simplifies to $\sqrt{10}=\sqrt{10}$. Since $\sqrt{10}=\sqrt{10}$, $x=5$ is a valid solution.

Conclusion

In conclusion, the solutions to the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$ are $x=-6$ and $x=5$. Therefore, the correct answer is D. $x=-6, x=5$.

Final Answer

The final answer is D. $x=-6, x=5$.

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Introduction

In our previous article, we explored the solution to the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$. We found that the solutions to the equation are $x=-6$ and $x=5$. In this article, we will answer some frequently asked questions about the solutions to the equation.

Q: What is the range of the solutions to the equation?

A: The range of the solutions to the equation is all real numbers.

Q: How did you find the solutions to the equation?

A: We found the solutions to the equation by first simplifying the expressions inside the square roots. We then canceled out the common factor $\sqrt{x+5}$ from both sides of the equation. After that, we squared both sides of the equation to solve for $x$.

Q: Why did you square both sides of the equation?

A: We squared both sides of the equation to eliminate the square root. This allowed us to solve for $x$.

Q: What if the equation had been $\sqrt{x^2+2x-25}=-\sqrt{x+5}$?

A: If the equation had been $\sqrt{x^2+2x-25}=-\sqrt{x+5}$, we would have had to consider the case where $x+5 \leq 0$. In this case, the equation would have become $\sqrt{x^2+2x-25}=-\sqrt{x+5}$, which would have led to the solutions $x=-6$ and $x=5$.

Q: Can you explain why $x=6$ is not a valid solution?

A: $x=6$ is not a valid solution because when we substitute $x=6$ into the original equation, we get $\sqrt{6^2+2(6)-25}=\sqrt{6+5}$, which simplifies to $\sqrt{36+12-25}=\sqrt{11}$, and further simplifies to $\sqrt{23}=\sqrt{11}$. Since $\sqrt{23} \neq \sqrt{11}$, $x=6$ is not a valid solution.

Q: Can you explain why $x=-6$ and $x=5$ are valid solutions?

A: $x=-6$ and $x=5$ are valid solutions because when we substitute $x=-6$ and $x=5$ into the original equation, we get $\sqrt{(-6)^2+2(-6)-25}=\sqrt{-6+5}$ and $\sqrt{5^2+2(5)-25}=\sqrt{5+5}$, respectively. These equations simplify to $\sqrt{36-12-25}=\sqrt{-1}$ and $\sqrt{25+10-25}=\sqrt{10}$, respectively. Since $\sqrt{(-1)}=\sqrt{-1}$ and $\sqrt{10}=\sqrt{10}$, $x=-6$ and $x=5$ are valid solutions.

Q: Can you provide more examples of equations that involve square roots?

A: Yes, here are a few examples of equations that involve square roots:

• $\sqrt{x^2-4x+4}=\sqrt{x-2}$
• $\sqrt{x^2+6x+9}=\sqrt{x+3}$
• $\sqrt{x^2-2x-15}=\sqrt{x-5}$

Q: How do you solve equations that involve square roots?

A: To solve equations that involve square roots, you can start by simplifying the expressions inside the square roots. You can then cancel out any common factors and square both sides of the equation to solve for $x$.

Conclusion

In conclusion, we have answered some frequently asked questions about the solutions to the equation $\sqrt{x^2+2x-25}=\sqrt{x+5}$. We have also provided some examples of equations that involve square roots and explained how to solve them.

Final Answer

The final answer is D. $x=-6, x=5$.