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

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Question

Solve for .

$\sqrt{x-2}=1$

Answer

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

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

Next, add two to both sides of the equation.

_____________

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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Question

Solve for .

$\sqrt{x}+4=3$

Answer

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.

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

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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Question

Solve for .

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

Answer

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.

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

Now, subtract one from both sides to solve for .

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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Question

Solve for .

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

Answer

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.

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

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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Question

Solve for .

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

Answer

To solve for , first add two to both sides.

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

____________

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

From here, multiply both sides by .

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

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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Question

Solve for .

$\sqrt{5-x}=2$

Answer

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

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

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.

_____________

From here, divide by negative one on both sides.

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

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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Question

Solve for .

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

Answer

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 .

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Question

Solve for .

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

Answer

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.

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

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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Question

Solve for .

$\sqrt{2-x}=7+3$

Answer

To solve for , combine the like terms on the right-hand side.

$\sqrt{2-x}=7+3$

$\sqrt{2-x}=10$

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

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

Now, subtract two from both sides so that all constants are on one side and all variables are on the other side.

_________________

Next, divide both sides by negative one.

$\frac{-x}{-1}=\frac{98}{-1}$

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

$\sqrt{2-x}=7+3$

$\\sqrt{2--98}=7+3 \\sqrt{100}=10 \\pm10=10$

Therefore, the solution is verified.

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Question

Solve for .

$2\sqrt{x}=14$

Answer

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.

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

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

$2\sqrt{x}=14$

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

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Question

Solve for .

Answer

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

______________

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.

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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Question

Solve for .

$2-\sqrt{x+1}=5+3$

Answer

To solve for , first combine the like terms on the right-hand side of the equation.

$2-\sqrt{x+1}=5+3$

$2-\sqrt{x+1}=8$

Next, subtract two from both sides.

$2-\sqrt{x+1}=8$

_______

$-\sqrt{x+1}=6$

Now, divide both sides by negative one.

$\frac{-\sqrt{x+1}}{-1}=\frac{6}{-1}$

$\sqrt{x+1}=-6$

From here, square both sides.

$(\sqrt{x+1})^2=(-6)^2$

Next, subtract one from both side to solve .

Lastly, check for extraneous solutions.

$2-\sqrt{x+1}=5+3$

$\2-\sqrt{35+1}=8 \2-\sqrt{36}=8 \2-\pm6=8 \2--6=8 \2-+6=-4 eq8$

Since one of the values is a correct solution this verifies as a solution.

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Question

Solve for .

$\sqrt{x-2}=1$

Answer

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

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

Next, add two to both sides of the equation.

_____________

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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Question

Solve for .

$\sqrt{x}+4=3$

Answer

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.

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

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.

Compare your answer with the correct one above

Question

Solve for .

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

Answer

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.

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

Now, subtract one from both sides to solve for .

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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Question

Solve for .

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

Answer

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.

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

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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Question

Solve for .

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

Answer

To solve for , first add two to both sides.

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

____________

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

From here, multiply both sides by .

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

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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Question

Solve for .

$\sqrt{5-x}=2$

Answer

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

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

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.

_____________

From here, divide by negative one on both sides.

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

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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Question

Solve for .

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

Answer

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 .

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Question

Solve for .

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

Answer

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.

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

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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