Solve 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. ( 2 X + 1 ) 1 3 = 1 (2x+1)^{\frac{1}{3}}=1 ( 2 X + 1 ) 3 1 ​ = 1

Introduction

Solving equations with radicals and exponents can be a challenging task, especially when dealing with extraneous solutions. In this article, we will explore how to solve and check for extraneous solutions in various types of equations, including those with square roots, cube roots, and fractional exponents.

Solving Equations with Square Roots

a. $\sqrt{x+7}=x-5$

To solve this equation, we start by isolating the square root expression. We can do this by squaring both sides of the equation.

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

This simplifies to:

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

Next, we can rearrange the equation to form a quadratic equation:

x^2 - 11x + 18 = 0

We can factor this quadratic equation to find its solutions:

(x - 6)(x - 3) = 0

This gives us two possible solutions: $x = 6$ and $x = 3$. However, we need to check if these solutions are valid by plugging them back into the original equation.

For $x = 6$, we have:

\sqrt{6+7} = 6-5
\sqrt{13} = 1

This is not true, so $x = 6$ is an extraneous solution.

For $x = 3$, we have:

\sqrt{3+7} = 3-5
\sqrt{10} = -2

This is also not true, so $x = 3$ is an extraneous solution.

Therefore, there are no valid solutions to this equation.

b. $\sqrt{2x-5}=7$

To solve this equation, we can square both sides to eliminate the square root:

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

This simplifies to:

2x - 5 = 49

Next, we can solve for $x$:

2x = 54
x = 27

We can check this solution by plugging it back into the original equation:

\sqrt{2(27)-5} = 7
\sqrt{49} = 7

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

c. $\sqrt{x+2}=10-x$

To solve this equation, we can square both sides to eliminate the square root:

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

This simplifies to:

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

Next, we can rearrange the equation to form a quadratic equation:

x^2 - 19x + 98 = 0

We can factor this quadratic equation to find its solutions:

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

This gives us two possible solutions: $x = 14$ and $x = 7$. However, we need to check if these solutions are valid by plugging them back into the original equation.

For $x = 14$, we have:

\sqrt{14+2} = 10-14
\sqrt{16} = -4

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

For $x = 7$, we have:

\sqrt{7+2} = 10-7
\sqrt{9} = 3

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

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

To solve this equation, we can isolate the expression with the fractional exponent:

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

Next, we can raise both sides to the power of $\frac{3}{2}$ to eliminate the fractional exponent:

(x-2) = 9^{\left(\frac{3}{2}\right)}
(x-2) = 27

Finally, we can solve for $x$:

x = 29

We can check this solution by plugging it back into the original equation:

(29-2)^{\left(\frac{2}{3}\right)} - 4 = 5
27^{\left(\frac{2}{3}\right)} - 4 = 5
9 - 4 = 5
5 = 5

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

e. $\sqrt{7x-6}-\sqrt{5x+2}=0$

To solve this equation, we can isolate one of the square root expressions:

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

Next, we can square both sides to eliminate the square roots:

7x - 6 = 5x + 2

This simplifies to:

2x = 8
x = 4

We can check this solution by plugging it back into the original equation:

\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.

f. $(2x+1)^{\frac{1}{3}}=1$

To solve this equation, we can raise both sides to the power of 3 to eliminate the fractional exponent:

(2x+1) = 1^3
(2x+1) = 1

Next, we can solve for $x$:

2x = 0
x = 0

We can check this solution by plugging it back into the original equation:

(2(0)+1)^{\frac{1}{3}} = 1
1^{\frac{1}{3}} = 1
1 = 1

This is true, so $x = 0$ is a valid solution.

Conclusion

Q: What is an extraneous solution?

A: An extraneous solution is a solution to an equation that is not valid or does not satisfy the original equation. In other words, it is a solution that is not a true solution to the equation.

Q: How do I know if a solution is extraneous?

A: To determine if a solution is extraneous, you need to plug it back into the original equation and check if it is true. If the solution does not satisfy the original equation, then it is an extraneous solution.

Q: Why do I need to check for extraneous solutions?

A: Checking for extraneous solutions is important because it ensures that the solutions you find are valid and accurate. If you don't check for extraneous solutions, you may end up with incorrect or invalid solutions.

Q: What are some common mistakes to avoid when checking for extraneous solutions?

A: Some common mistakes to avoid when checking for extraneous solutions include:

• Not plugging the solution back into the original equation
• Not checking if the solution satisfies the original equation
• Not considering the possibility of extraneous solutions
• Not being careful when simplifying expressions or solving equations

Q: How do I check for extraneous solutions in equations with radicals?

A: To check for extraneous solutions in equations with radicals, you need to square both sides of the equation to eliminate the radical. Then, you can plug the solution back into the original equation to check if it is true.

Q: How do I check for extraneous solutions in equations with fractional exponents?

A: To check for extraneous solutions in equations with fractional exponents, you need to raise both sides of the equation to the power of the reciprocal of the exponent. Then, you can plug the solution back into the original equation to check if it is true.

Q: What are some tips for solving and checking for extraneous solutions?

A: Some tips for solving and checking for extraneous solutions include:

• Be careful when simplifying expressions or solving equations
• Check your work carefully to avoid mistakes
• Consider the possibility of extraneous solutions
• Plug the solution back into the original equation to check if it is true
• Use a calculator or computer software to check your work if necessary

Q: How do I know if I have found all the solutions to an equation?

A: To determine if you have found all the solutions to an equation, you need to check if the equation is true for all possible values of the variable. If the equation is not true for all possible values of the variable, then you may have missed some solutions.

Q: What are some common types of equations that require checking for extraneous solutions?

A: Some common types of equations that require checking for extraneous solutions include:

• Equations with radicals
• Equations with fractional exponents
• Equations with absolute values
• Equations with square roots
• Equations with cube roots

Conclusion

Checking for extraneous solutions is an important step in solving equations, especially those with radicals and fractional exponents. By following the tips and techniques outlined in this article, you can ensure that you find all the valid solutions to an equation and avoid extraneous solutions. Remember to always check your work carefully and consider the possibility of extraneous solutions.