We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

New SAT Math - Calculator : Inequalities and Absolute Value

Study concepts, example questions & explanations for New SAT Math - Calculator

Create An Account

All New SAT Math - Calculator Resources

2 Diagnostic Tests 27 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Systems Of Inequalities

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

$\sqrt{x + 7} > x + 1$

Possible Answers:

x = 2.0

2.0 < x < -3.0

2.0 < x

$-3.0 \leq x$

-3.0 < x < 2.0

Correct answer:

-3.0 < x < 2.0

Explanation:

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

$\left( \sqrt{x + 7} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 7 > x^{2} + 2 x + 1$

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

$x + 7 -\left( x + 7 \right)> x^{2} + 2 x + 1 -\left( x + 7 \right)$

$0 > x^{2} + x - 6$

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 = -6 .

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

x = 2.0

x = -3.0

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.

-3.0 < x < 2.0

Report an Error

Example Question #502 : New Sat

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

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

Possible Answers:

x = 2.5

$-3.5 \leq x$

-3.5 < x < 2.5

2.5 < x < -3.5

2.5 < 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 #1 : Inequalities And Absolute Value

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

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

Possible Answers:

0.0 < x

x = 0.0

0.0 < x < -1.0

-1.0 < x < 0.0

$-1.0 \leq x$

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 #3 : Inequalities And Absolute Value

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

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

Possible Answers:

$-1.0 \leq x$

0.0 < x < -1.0

0.0 < x

x = 0.0

-1.0 < x < 0.0

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 #511 : New Sat

$\frac{3x+4}{4}<\frac{x}{2}$

Which of the following provides the complete solution set for given the above inequality?

Possible Answers:

x<-2

$x<\frac{1}{2}$

x>4

x<2

x<-4

Correct answer:

x<-4

Explanation:

To solve this problem, first cross-multiply the inequality to eliminate the denominators. Note that while this is an inequality, you can safely multiply by both denominators since both are positive so there is no need to consider flipping the direction of the inequality. The result of this step is:

6x+8<4x

Then you can combine like terms by subtracting from both sides:

2x+8<0

Then to isolate the variable term, subtract from both sides:

2x<-8

Finally, divide both sides by to get the variable alone:

x<-4

Report an Error

Example Question #1 : Systems Of Inequalities

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

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

Possible Answers:

x = 1.0

-2.0 < x < 1.0

1.0 < x < -2.0

1.0 < x

$-2.0 \leq x$

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:

$-1.0 \leq x$

-1.0 < x < 0.0

x = 0.0

0.0 < x < -1.0

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 #2 : Systems Of Inequalities

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

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

Possible Answers:

$-2.8 \leq x$

-2.8 < x < 1.8

1.8 < x

x = 1.8

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 #121 : New Sat Math Calculator

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

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

Possible Answers:

x = 5.4

-6.4 < x < 5.4

5.4 < x

5.4 < x < -6.4

$-6.4 \leq x$

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 #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.

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

Possible Answers:

x = 3.5

3.5 < x

$-4.5 \leq x$

3.5 < x < -4.5

-4.5 < x < 3.5

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

View Tutors

Evgenii
Certified Tutor

View Tutors

Stephanie
Certified Tutor

Boise State University, Bachelor in Arts, Social Work. Arizona State University, Master of Arts, Social Work.

View Tutors

Gavin
Certified Tutor

The University of Texas at San Antonio, Bachelor of Science, Civil Engineering.

All New SAT Math - Calculator Resources

2 Diagnostic Tests 27 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in Philadelphia, GMAT Tutors in Phoenix, Reading Tutors in New York City, English Tutors in Houston, ISEE Tutors in Washington DC, Algebra Tutors in Atlanta, LSAT Tutors in San Francisco-Bay Area, SAT Tutors in Dallas Fort Worth, ACT Tutors in Seattle, GRE Tutors in Washington DC

Popular Courses & Classes

GMAT Courses & Classes in Phoenix, SAT Courses & Classes in New York City, SAT Courses & Classes in San Diego, ACT Courses & Classes in Boston, GMAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Philadelphia, SSAT Courses & Classes in Miami, ISEE Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Chicago, ACT Courses & Classes in Miami

Popular Test Prep

MCAT Test Prep in Washington DC, ACT Test Prep in Boston, GRE Test Prep in San Diego, GRE Test Prep in Chicago, SAT Test Prep in Boston, ISEE Test Prep in Phoenix, ISEE Test Prep in New York City, SAT Test Prep in Los Angeles, SSAT Test Prep in New York City, GRE Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

New SAT Math - Calculator : Inequalities and Absolute Value

Study concepts, example questions & explanations for New SAT Math - Calculator

All New SAT Math - Calculator Resources

Example Questions

Example Question #1 : Systems Of Inequalities

Example Question #502 : New Sat

Example Question #1 : Inequalities And Absolute Value

Example Question #3 : Inequalities And Absolute Value

Example Question #511 : New Sat

Example Question #1 : Systems Of Inequalities

Example Question #1 : Systems Of Inequalities

Example Question #2 : Systems Of Inequalities

Example Question #121 : New Sat Math Calculator

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

All New SAT Math - Calculator Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: