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Common Core: High School - Algebra : Solve Simple Rational and Radical One Variable Equations, Show Extraneous Solutions: CCSS.Math.Content.HSA-REI.A.2

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Example Question #1 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$\sqrt{x-2}=1$

Possible Answers:

x=2

x=3

x=-1

x=-3

x=1

Correct answer:

x=3

Explanation:

To solve for , square both terms to cancel out the square root sign on the left-hand side.

$(\sqrt{x-2})^2=(1)^2$

x-2=1

Next, add two to both sides of the equation.

x-2=1

_____________

x=3

From here, check for extraneous solutions by substituting the value found for into the original equation.

$\sqrt{x-2}=1$

$\\ \sqrt{3-2}=1 \\\sqrt{1}=1 \\\pm1=1$

Since the square root of one is either positive or negative one, the solution found is verified.

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Example Question #2 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$1-\sqrt{x+1}=5$

Possible Answers:

x=24

x=5

x=16

x=15

x=4

Correct answer:

x=15

Explanation:

To solve for , first subtract one from both sides.

$1-\sqrt{x+1}=5$

____________________

$-\sqrt{x+1}=4$

From here, divide both sides by negative one.

$-\frac{\sqrt{x+1}}{-1}=\frac{4}{-1}$

$\sqrt{x+1}=-4$

Next, square both sides to cancel the square root sign on the left-hand side.

$(\sqrt{x+1})^2=(-4)^2$

x+1=16

Now, subtract one from both sides to solve for .

x+1=16

____________

x=15

Lastly, check for extraneous solutions by substituting the value found for into the original equation.

$1-\sqrt{x+1}=5$

$\\1-\sqrt{15+1}=5 \\1-\sqrt{16}=5 \\1-\pm4=5$

Thus, the answer is verified.

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Example Question #3 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$\frac{1}{x}-2=10$

Possible Answers:

x=12

x=-12

$x=\frac{1}{12}$

$x=-\frac{1}{12}$

$x=\frac{1}{8}$

Correct answer:

$x=\frac{1}{12}$

Explanation:

To solve for , add two to both sides of the equation.

$\frac{1}{x}-2=10$

______________

$\frac{1}{x}=12$

Now, multiply by on both sides. This will move the variable from the denominator on one side to the numerator of the other side.

$x\cdot \frac{1}{x}=12\cdot x$

1=12x

Lastly, divide both sides by twelve.

$\frac{1}{12}=\frac{12x}{12}$

The twelve in the numerator and the twelve in the denominator cancel out thus, solving for .

$\frac{1}{12}=x$

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Example Question #4 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$\sqrt{x}+4=3$

Possible Answers:

x=3

x=5

x=1

x=2

x=-1

Correct answer:

x=1

Explanation:

To solve for , first subtract four from both sides so all constants are on one side.

$\sqrt{x}+4=3$

_____________

$\sqrt{x}=-1$

Now, square both sides to cancel the square root sign on the left-hand side.

$(\sqrt{x})^2=(-1)^2$

x=1

Recall that when a negative number is squared the result is always a positive value.

From here, check for extraneous solutions by substituting in the value found for .

$\\\sqrt{x}+4=3 \\\sqrt{1}+4=3 \\\pm1+4=3 \\-1+4=3$

Since the square root of a value results in a positive and a negative value, our solution is verified.

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Example Question #5 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$\sqrt{\frac{x}{2}}=6$

Possible Answers:

x=36

x=24

x=6

x=72

x=-72

Correct answer:

x=72

Explanation:

To solve for start by squaring both sides to eliminate the square root sign.

$\left( \sqrt{\frac{x}{2}}\right)^2=(6)^2$

$\frac{x}{2}=36$

From here multiply both sides by two. On the left-hand side, the two in the numerator will cancel out the two in the denominator.

$2\cdot \frac{x}{2}=36\cdot 2$

x=72

From here, check for extraneous solutions by substituting in the value found for into the original equation.

$\sqrt{\frac{x}{2}}=6$

$\sqrt{\frac{72}{2}}=6$

$\sqrt{36}=6$

$\pm6=6$

Thus, the solution is verified.

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Example Question #6 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$\frac{3}{x}-2=5$

Possible Answers:

$x=\frac{7}{3}$

$x=\frac{3}{7}$

x=21

$x=-\frac{3}{7}$

$x=-\frac{7}{3}$

Correct answer:

$x=\frac{3}{7}$

Explanation:

To solve for , first add two to both sides.

$\frac{3}{x}-2=5$

____________

$\frac{3}{x}=7$

From here, multiply both sides by .

$x\cdot \frac{3}{x}=7\cdot x$

3=7x

Now divide by seven to solve for .

$\frac{3}{7}=\frac{7x}{7}$

The seven in the numerator and seven in the denominator cancel out, thus solving for .

$\frac{3}{7}=x$

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Example Question #7 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$\sqrt{5-x}=2$

Possible Answers:

x=0

x=-1

x=2

x=1

x=-2

Correct answer:

x=1

Explanation:

To solve for , first square both sides of the equation. Squaring a square root sign will cancel them out.

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

5-x=4

Now, subtract five from both sides to get all constants on one side of the equation while all variables are on the other side of the equation.

5-x=4

_____________

-x=-1

From here, divide by negative one on both sides.

$\frac{-x}{-1}=\frac{-1}{-1}$

x=1

Lastly, check for extraneous solutions by substituting in the value found for into the original equation.

$\\\sqrt{5-1}=2 \\\sqrt{4}=2 \\\pm2=2$

Thus, the solution is verified.

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Example Question #1 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$\frac{x}{2}=8$

Possible Answers:

x=8

x=6

x=2

x=16

x=4

Correct answer:

x=16

Explanation:

To solve for , simply multiply both sides of the equation by two.

$\frac{x}{2}=8$

$2\cdot \frac{x}{2}=8\cdot 2$

The two in the numerator cancels the two in the denominator, thus solving for .

x=16

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Example Question #9 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

$2\sqrt{x}=14$

Possible Answers:

x=49

x=-49

x=-14

x=14

x=7

Correct answer:

x=49

Explanation:

To solve for , first divide by two.

$2\sqrt{x}=14$

$\frac{2\sqrt{x}}{2}=\frac{14}{2}$

The two in the denominator cancels the two in the numerator.

$\sqrt{x}=7$

From here, square both sides of the equation. Squaring a square root cancels it out.

$(\sqrt{x})^2=(7)^2$

x=49

To check for extraneous solutions, simply substitute into the original equation the value found for .

$2\sqrt{x}=14$

$\\2\sqrt{49}=14 \\2\cdot \pm 7=14 \\2(7)=14 \\2(-7)=-14\neq 14$

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Example Question #10 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Solve for .

x^2-3=13

Possible Answers:

x=4

x=16

$x=\pm4$

$x=\pm16$

x=-4

Correct answer:

$x=\pm4$

Explanation:

To solve for , add three to both sides of the equation.

x^2-3=13

______________

x^2=16

From here, take the square root of both sides. Taking the square root of a squared term eliminates the square.

$\sqrt{x^2}=\sqrt{16}$

$x=\pm4$

Next, check for extraneous solutions. Substitute each potential solution into the original equation to verify if it results in a legal solution.

x^2-3=13

Substituting in positive four.

$\\4^2-3=13 \\16-3=13 \\13=13$

Substituting in negative four.

$\\(-4)^2-3=13 \\16-3=13 \\13=13$

Therefore, both answers are verified solutions.

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Common Core: High School - Algebra : Solve Simple Rational and Radical One Variable Equations, Show Extraneous Solutions: CCSS.Math.Content.HSA-REI.A.2

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #1 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #2 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #3 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #4 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #5 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #6 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #7 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #1 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #9 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

Example Question #10 : Solve Simple Rational And Radical One Variable Equations, Show Extraneous Solutions: Ccss.Math.Content.Hsa Rei.A.2

All Common Core: High School - Algebra Resources

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