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Common Core: High School - Algebra : Solve One Variable Linear Equations and Inequalities: CCSS.Math.Content.HSA-REI.B.3

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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

Report an Error

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

Example Question #7 : 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.

$\sqrt{x + 35} > x + 1$

Possible Answers:

$-6.4 \leq x$

x = 5.4

5.4 < x < -6.4

5.4 < x

-6.4 < x < 5.4

Correct answer:

-6.4 < x < 5.4

Explanation:

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

$\left( \sqrt{x + 35} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 35 > x^{2} + 2 x + 1$

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

$x + 35 -\left( x + 35 \right)> x^{2} + 2 x + 1 -\left( x + 35 \right)$

$0 > x^{2} + x - 34$

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 = -34 .

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

x = 5.4

x = -6.4

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.

-6.4 < x < 5.4

Report an Error

Example Question #681 : Algebra

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

$\sqrt{x + 17} > x + 1$

Possible Answers:

x = 3.5

3.5 < x < -4.5

-4.5 < x < 3.5

3.5 < x

$-4.5 \leq x$

Correct answer:

-4.5 < x < 3.5

Explanation:

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

$\left( \sqrt{x + 17} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 17 > x^{2} + 2 x + 1$

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

$x + 17 -\left( x + 17 \right)> x^{2} + 2 x + 1 -\left( x + 17 \right)$

$0 > x^{2} + x - 16$

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 = -16 .

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

x = 3.5

x = -4.5

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.

-4.5 < x < 3.5

Report an Error

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.

$\sqrt{x + 8} > x + 1$

Possible Answers:

$-3.2 \leq x$

x = 2.2

2.2 < x

2.2 < x < -3.2

-3.2 < x < 2.2

Correct answer:

-3.2 < x < 2.2

Explanation:

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

$\left( \sqrt{x + 8} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 8 > x^{2} + 2 x + 1$

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

$x + 8 -\left( x + 8 \right)> x^{2} + 2 x + 1 -\left( x + 8 \right)$

$0 > x^{2} + x - 7$

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 = -7 .

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

x = 2.2

x = -3.2

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.

-3.2 < x < 2.2

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.

$\sqrt{x + 15} > x + 1$

Possible Answers:

x = 3.3

-4.3 < x < 3.3

3.3 < x

$-4.3 \leq x$

3.3 < x < -4.3

Correct answer:

-4.3 < x < 3.3

Explanation:

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

$\left( \sqrt{x + 15} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 15 > x^{2} + 2 x + 1$

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

$x + 15 -\left( x + 15 \right)> x^{2} + 2 x + 1 -\left( x + 15 \right)$

$0 > x^{2} + x - 14$

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 = -14 .

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

x = 3.3

x = -4.3

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.

-4.3 < x < 3.3

Report an Error

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Common Core: High School - Algebra : Solve One Variable Linear Equations and Inequalities: CCSS.Math.Content.HSA-REI.B.3

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

All Common Core: High School - Algebra Resources

Example Questions

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

Example Question #7 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #681 : Algebra

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

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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