Solve Each Equation And Check For Extraneous Solutions.a. X + 7 = X − 5 \sqrt{x+7} = X - 5 X + 7 ​ = X − 5 B. 2 X − 5 = 7 \sqrt{2x-5} = 7 2 X − 5 ​ = 7 C. X + 2 = 10 − X \sqrt{x+2} = 10 - X X + 2 ​ = 10 − X D. ( X − 2 ) ( 2 3 ) − 4 = 5 (x-2)^{\left(\frac{2}{3}\right)} - 4 = 5 ( X − 2 ) ( 3 2 ​ ) − 4 = 5 E. 7 X − 6 − 5 X + 2 = 0 \sqrt{7x-6} - \sqrt{5x+2} = 0 7 X − 6 ​ − 5 X + 2 ​ = 0 F.

===========================================================

Introduction

---

Solving equations with square roots can be a challenging task, especially when dealing with extraneous solutions. In this article, we will explore how to solve each of the given equations and check for extraneous solutions. We will use algebraic techniques to isolate the variable and then verify the solutions.

Solving Equation a: $\sqrt{x+7} = x - 5$

---

To solve this equation, we will start by isolating the square root expression.

Step 1: Square Both Sides

---

We will square both sides of the equation to eliminate the square root.

$$\left(\sqrt{x+7}\right)^2 = (x-5)^2$$

This simplifies to:

$$x+7 = x^2 - 10x + 25$$

Step 2: Rearrange the Equation

---

We will rearrange the equation to form a quadratic equation.

$$x^2 - 11x + 18 = 0$$

Step 3: Factor the Quadratic Equation

---

We will factor the quadratic equation to find the solutions.

$$(x-9)(x-2) = 0$$

This gives us two possible solutions:

$$x = 9 \text{ or } x = 2$$

Step 4: Check for Extraneous Solutions

---

We will substitute each solution back into the original equation to check for extraneous solutions.

For $x = 9$:

$$\sqrt{9+7} = 9 - 5$$

$$\sqrt{16} = 4$$

This is true, so $x = 9$ is a valid solution.

For $x = 2$:

$$\sqrt{2+7} = 2 - 5$$

$$\sqrt{9} = -3$$

This is not true, so $x = 2$ is an extraneous solution.

Solving Equation b: $\sqrt{2x-5} = 7$

---

To solve this equation, we will start by isolating the square root expression.

Step 1: Square Both Sides

---

We will square both sides of the equation to eliminate the square root.

$$\left(\sqrt{2x-5}\right)^2 = 7^2$$

This simplifies to:

$$2x-5 = 49$$

Step 2: Rearrange the Equation

---

We will rearrange the equation to solve for $x$.

$$2x = 54$$

$$x = 27$$

Step 3: Check for Extraneous Solutions

---

We will substitute the solution back into the original equation to check for extraneous solutions.

$$\sqrt{2(27)-5} = 7$$

$$\sqrt{49} = 7$$

This is true, so $x = 27$ is a valid solution.

Solving Equation c: $\sqrt{x+2} = 10 - x$

---

To solve this equation, we will start by isolating the square root expression.

Step 1: Square Both Sides

---

We will square both sides of the equation to eliminate the square root.

$$\left(\sqrt{x+2}\right)^2 = (10-x)^2$$

This simplifies to:

$$x+2 = 100 - 20x + x^2$$

Step 2: Rearrange the Equation

---

We will rearrange the equation to form a quadratic equation.

$$x^2 - 21x + 98 = 0$$

Step 3: Factor the Quadratic Equation

---

We will factor the quadratic equation to find the solutions.

$$(x-14)(x-7) = 0$$

This gives us two possible solutions:

$$x = 14 \text{ or } x = 7$$

Step 4: Check for Extraneous Solutions

---

We will substitute each solution back into the original equation to check for extraneous solutions.

For $x = 14$:

$$\sqrt{14+2} = 10 - 14$$

$$\sqrt{16} = -4$$

This is not true, so $x = 14$ is an extraneous solution.

For $x = 7$:

$$\sqrt{7+2} = 10 - 7$$

$$\sqrt{9} = 3$$

This is true, so $x = 7$ is a valid solution.

Solving Equation d: $(x-2)^{\left(\frac{2}{3}\right)} - 4 = 5$

---

To solve this equation, we will start by isolating the expression with the exponent.

Step 1: Add 4 to Both Sides

---

We will add 4 to both sides of the equation to isolate the expression with the exponent.

$$(x-2)^{\left(\frac{2}{3}\right)} = 9$$

Step 2: Raise Both Sides to the Power of $\frac{3}{2}$

---

We will raise both sides of the equation to the power of $\frac{3}{2}$ to eliminate the exponent.

$$(x-2) = 27$$

Step 3: Solve for $x$

---

We will solve for $x$.

$$x = 29$$

Step 4: Check for Extraneous Solutions

---

We will substitute the solution back into the original equation to check for extraneous solutions.

$$(29-2)^{\left(\frac{2}{3}\right)} - 4 = 5$$

$$27^{\left(\frac{2}{3}\right)} - 4 = 5$$

$$9 - 4 = 5$$

This is true, so $x = 29$ is a valid solution.

Solving Equation e: $\sqrt{7x-6} - \sqrt{5x+2} = 0$

---

To solve this equation, we will start by isolating the square root expressions.

Step 1: Add $\sqrt{5x+2}$ to Both Sides

---

We will add $\sqrt{5x+2}$ to both sides of the equation to isolate the square root expressions.

$$\sqrt{7x-6} = \sqrt{5x+2}$$

Step 2: Square Both Sides

---

We will square both sides of the equation to eliminate the square root expressions.

$$(\sqrt{7x-6})^2 = (\sqrt{5x+2})^2$$

This simplifies to:

$$7x-6 = 5x+2$$

Step 3: Rearrange the Equation

---

We will rearrange the equation to solve for $x$.

$$2x = 8$$

$$x = 4$$

Step 4: Check for Extraneous Solutions

---

We will substitute the solution back into the original equation to check for extraneous solutions.

$$\sqrt{7(4)-6} - \sqrt{5(4)+2} = 0$$

$$\sqrt{26} - \sqrt{22} = 0$$

This is not true, so $x = 4$ is an extraneous solution.

Conclusion

---

In this article, we have solved each of the given equations and checked for extraneous solutions. We have used algebraic techniques to isolate the variable and then verified the solutions. We have also discussed the importance of checking for extraneous solutions in solving equations with square roots.

=============================================

Introduction

---

Solving equations with square roots can be a challenging task, especially when dealing with extraneous solutions. In this article, we will answer some of the most frequently asked questions about solving equations with square roots.

Q: What is an extraneous solution?

---

A: An extraneous solution is a solution that is not valid for the original equation. It is a solution that is introduced during the process of solving the equation, but it does not satisfy the original equation.

Q: How do I check for extraneous solutions?

---

A: To check for extraneous solutions, you need to substitute the solution back into the original equation and verify that it is true. If the solution does not satisfy the original equation, then it is an extraneous solution.

Q: What are some common mistakes to avoid when solving equations with square roots?

---

A: Some common mistakes to avoid when solving equations with square roots include:

• Not checking for extraneous solutions
• Not squaring both sides of the equation correctly
• Not rearranging the equation correctly
• Not factoring the quadratic equation correctly

Q: How do I solve an equation with a square root on both sides?

---

A: To solve an equation with a square root on both sides, you need to isolate the square root expressions and then square both sides of the equation. This will eliminate the square root expressions and allow you to solve for the variable.

Q: What is the difference between a square root and an exponent?

---

A: A square root is a mathematical operation that finds the number that, when multiplied by itself, gives a specified value. An exponent, on the other hand, is a mathematical operation that raises a number to a power. For example, $\sqrt{16} = 4$ and $4^2 = 16$.

Q: How do I solve an equation with a fractional exponent?

---

A: To solve an equation with a fractional exponent, you need to raise both sides of the equation to the power of the reciprocal of the exponent. For example, if the equation is $(x-2)^{\left(\frac{2}{3}\right)} = 9$, you would raise both sides of the equation to the power of $\frac{3}{2}$ to eliminate the exponent.

Q: What are some real-world applications of solving equations with square roots?

---

A: Some real-world applications of solving equations with square roots include:

• Calculating the length of a shadow or the height of a building
• Determining the area of a square or rectangle
• Finding the volume of a cube or rectangular prism
• Solving problems in physics, engineering, and other fields that involve square roots

Q: How do I know if an equation has a solution?

---

A: To determine if an equation has a solution, you need to check if the equation is true for any value of the variable. If the equation is true for any value of the variable, then it has a solution. If the equation is not true for any value of the variable, then it does not have a solution.

Q: What are some common types of equations that involve square roots?

---

A: Some common types of equations that involve square roots include:

• Linear equations with square roots
• Quadratic equations with square roots
• Exponential equations with square roots
• Trigonometric equations with square roots

Conclusion

---

In this article, we have answered some of the most frequently asked questions about solving equations with square roots. We have discussed the importance of checking for extraneous solutions, common mistakes to avoid, and real-world applications of solving equations with square roots. We hope that this article has been helpful in answering your questions and providing you with a better understanding of solving equations with square roots.