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Common Core: High School - Algebra : Reasoning with Equations & Inequalities

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

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All Common Core: High School - Algebra Resources

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Example Questions

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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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Example Question #21 : Reasoning With Equations & Inequalities

Solve for .

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

Possible Answers:

x=102

x=-100

x=-98

x=-102

x=98

Correct answer:

x=-98

Explanation:

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.

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

2-x=100

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

2-x=100

_________________

-x=98

Next, divide both sides by negative one.

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

x=-98

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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Example Question #22 : Reasoning With Equations & Inequalities

Solve for .

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

Possible Answers:

x=36

x=35

x=-37

x=37

x=-35

Correct answer:

x=35

Explanation:

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.

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

x+1=36

Next, subtract one from both side to solve .

x+1=36

______________

x=35

Lastly, check for extraneous solutions.

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

$\\2-\sqrt{35+1}=8 \\2-\sqrt{36}=8 \\2-\pm6=8 \\2--6=8 \\2-+6=-4\neq8$

Since one of the values is a correct solution this verifies x=35 as a solution.

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Example Question #1 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 10} > x + 1$

Possible Answers:

2.5 < x

-3.5 < x < 2.5

2.5 < x < -3.5

x = 2.5

$-3.5 \leq x$

Correct answer:

-3.5 < x < 2.5

Explanation:

The first step is to square each side of the inequality.

$\left( \sqrt{x + 10} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 10 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 10 -\left( x + 10 \right)> x^{2} + 2 x + 1 -\left( x + 10 \right)$

$0 > x^{2} + x - 9$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 1 , and c = -9 .

Now plug these values into the quadratic equation, and we get.

x = 2.5

x = -3.5

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

-3.5 < x < 2.5

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Example Question #2 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 1} > x + 1$

Possible Answers:

$-1.0 \leq x$

x = 0.0

0.0 < x < -1.0

-1.0 < x < 0.0

0.0 < x

Correct answer:

-1.0 < x < 0.0

Explanation:

The first step is to square each side of the inequality.

$\left( \sqrt{x + 1} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 1 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 1 -\left( x + 1 \right)> x^{2} + 2 x + 1 -\left( x + 1 \right)$

$0 > x^{2} + x$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 1 , and c = 0 .

Now plug these values into the quadratic equation, and we get.

x = 0.0

x = -1.0

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

-1.0 < x < 0.0

Report an Error

Example Question #3 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 1} > x + 1$

Possible Answers:

x = 0.0

$-1.0 \leq x$

0.0 < x

-1.0 < x < 0.0

0.0 < x < -1.0

Correct answer:

-1.0 < x < 0.0

Explanation:

The first step is to square each side of the inequality.

$\left( \sqrt{x + 1} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 1 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 1 -\left( x + 1 \right)> x^{2} + 2 x + 1 -\left( x + 1 \right)$

$0 > x^{2} + x$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 1 , and c = 0 .

Now plug these values into the quadratic equation, and we get.

x = 0.0

x = -1.0

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

-1.0 < x < 0.0

Report an Error

Example Question #4 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 3} > x + 1$

Possible Answers:

1.0 < x

x = 1.0

-2.0 < x < 1.0

1.0 < x < -2.0

$-2.0 \leq x$

Correct answer:

-2.0 < x < 1.0

Explanation:

The first step is to square each side of the inequality.

$\left( \sqrt{x + 3} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 3 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 3 -\left( x + 3 \right)> x^{2} + 2 x + 1 -\left( x + 3 \right)$

$0 > x^{2} + x - 2$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 1 , and c = -2 .

Now plug these values into the quadratic equation, and we get.

x = 1.0

x = -2.0

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

-2.0 < x < 1.0

Report an Error

Example Question #1 : Systems Of Inequalities

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 1} > x + 1$

Possible Answers:

x = 0.0

-1.0 < x < 0.0

0.0 < x < -1.0

0.0 < x

$-1.0 \leq x$

Correct answer:

-1.0 < x < 0.0

Explanation:

The first step is to square each side of the inequality.

$\left( \sqrt{x + 1} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 1 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 1 -\left( x + 1 \right)> x^{2} + 2 x + 1 -\left( x + 1 \right)$

$0 > x^{2} + x$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 1 , and c = 0 .

Now plug these values into the quadratic equation, and we get.

x = 0.0

x = -1.0

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

-1.0 < x < 0.0

Report an Error

Example Question #1 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 6} > x + 1$

Possible Answers:

x = 1.8

-2.8 < x < 1.8

1.8 < x

$-2.8 \leq x$

1.8 < x < -2.8

Correct answer:

-2.8 < x < 1.8

Explanation:

The first step is to square each side of the inequality.

$\left( \sqrt{x + 6} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 6 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 6 -\left( x + 6 \right)> x^{2} + 2 x + 1 -\left( x + 6 \right)$

$0 > x^{2} + x - 5$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 1 , and c = -5 .

Now plug these values into the quadratic equation, and we get.

x = 1.8

x = -2.8

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

-2.8 < x < 1.8

Report an Error

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : Reasoning with Equations & Inequalities

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

All Common Core: High School - Algebra Resources

Example Questions

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

Example Question #21 : Reasoning With Equations & Inequalities

Example Question #22 : Reasoning With Equations & Inequalities

Example Question #1 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #2 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #3 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #4 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #1 : Systems Of Inequalities

Example Question #1 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

All Common Core: High School - Algebra Resources

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