Call Now to Set Up Tutoring:

(888) 888-0446

Algebra II : Solving Quadratic Equations

Study concepts, example questions & explanations for Algebra II

Create An Account

All Algebra II Resources

10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Quadratic Formula

Solve $-3x^2+6x-12{}$ .

Possible Answers:

$x=6;x=-\frac{2}{3}$

x=1;x=-1

x=-3

x=3;x=2

No real solutions

Correct answer:

No real solutions

Explanation:

-3x^2+6x-12=0

This function cannot be factor therefore, use the quadratic equation.

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

Since the original equation is in the form ax^2+bx+c where

a=-3, b=6, c=-12 .

Therefore,

$\frac{-6\pm\sqrt{6^2-4(-3)(-12)}}{2(-3)}$

$\frac{-6\pm\sqrt{36-144}}{-6}$

$\frac{-6\pm\sqrt{-108}}{-6}$

Since the value of the discriminant (the value beneath the square root symbol) is negative, this function has no real solutions.

Report an Error

Example Question #172 : Solving Quadratic Equations

Solve x^2+8x-2=0 .

Possible Answers:

$x={-4\pm {3\sqrt{2}}}$

$x={4\pm {3\sqrt{2}}}$

$x={-4\pm {3\sqrt{6}}}$

$x={-4\pm {6\sqrt{2}}}$

$x={-8\pm {3\sqrt{2}}}$

Correct answer:

$x={-4\pm {3\sqrt{2}}}$

Explanation:

x^2+8x-2=0

This particular function cannot be factored therefore, use the quadratic formula to solve.

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

Since the function is in the form ax^2+bx+c where

a=1,b=8, c=-2

the quadratic formula becomes as follows.

$\frac{-8\pm {\sqrt{8^2-4(1)(-2)}}}{2(1)}$

$\frac{-8\pm {\sqrt{64+8}}}{2}$

$\frac{-8\pm {\sqrt{72}}}{2}$

$\frac{-8\pm {\sqrt{9}\times\sqrt{4}\times\sqrt{2}}}{2}$

$\frac{-8\pm {3\times2\times\sqrt{2}}}{2}$

${-4\pm {3\sqrt{2}}}$

Report an Error

Example Question #1621 : Algebra Ii

Use the quadratic formula to solve for x:

4x+x^2+2

Possible Answers:

$-4\pm\sqrt{8}$

$-4\pm2\sqrt{2}$

$-4\pm\sqrt{2}$

$-2\pm\sqrt{2}$

${-2\pm2\sqrt{2}}$

Correct answer:

$-2\pm\sqrt{2}$

Explanation:

To solve this problem, you must first rewrite the equation into ax^2+bx+c form (quadratic form).

After this you plug the numbers into the following quadratic equation:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Which upon doing you get:

$x=\frac{-4\pm \sqrt{16-8}}{2}$

This simplifies to:

$x=-2\pm\sqrt{2}$

Report an Error

Example Question #174 : Solving Quadratic Equations

Use the quadratic formula to find the roots of the quadratic,

2x^2 + 3x-7 = 0

Possible Answers:

$x = \left\{\frac{3}{4} + \frac{\sqrt{65}}{2}, \frac{3}{4} - \frac{\sqrt{65}}{2} \right\}$

$x = \left\{-\frac{3}{4} + \frac{3\sqrt{6}}{4}i, -\frac{3}{4} - \frac{3\sqrt{6}}{4}i \right\}$

$x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right\}$

$x = \left\{-\frac{3}{4} + \frac{3\sqrt{6}}{4} -\frac{3}{4} - \frac{3\sqrt{6}}{4} \right\}$

$x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{2}i, -\frac{3}{4} - \frac{\sqrt{65}}{2}i \right\}$

Correct answer:

$x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right\}$

Explanation:

2x^2 +3x - 7 = 0

Recall the general form of a quadratic,

ax^2 +bx+c = 0

The solution set has the form,

$x = \frac{-b \pm \sqrt{b^2 - 4ca}}{2a}$

For our particular case, a = 2 , b = 3 , and c = -7

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-7)}}{2(2)}$

$x = \frac{-3 \pm \sqrt{65}}{4}$

$x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right\}$

Report an Error

Example Question #31 : Quadratic Formula

Use the quadratic formula to find the roots of the following equation.

2x^2-12x+16=0

Possible Answers:

x=12, x=8

x=-6,x=-12

x=-12,x=16

x=4,x=2

x=2,x=-4

Correct answer:

x=4,x=2

Explanation:

2x^2-12x+16=0

First, simplify the equation, so that the numbers are easier to work with. We can see that we can factor a 2 from each term.

2(x^2-6x+8)=0

Now we can divide both sides by 2, to further simplify.

$\frac{2}{2}(x^2-6x+8)=\frac{0}{2}$

x^2-6x+8 = 0

Now that we have simplified we can apply the quadratic formula.

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

where a, b, and c are the constants defined as follows:

ax^2+bx+c = 0

This means our a is 1, our b is -6, and our c is 8.

Finally, lets plug the numbers into the formula:

$\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(8)}}{2(1)}$

$\frac{6\pm \sqrt{36-32}}{2}$

$\frac{-6\pm \sqrt{4}}{2}$

$\frac{-6\pm 2}{2}$

$\frac{-4}{2}$ and $\frac{-8}{2}$

or more simply:

2, 4

These are the roots of the equation.

Report an Error

Example Question #171 : Solving Quadratic Equations

Find the roots of the equation using the quadratic equation.

4x^2-9x+3

Possible Answers:

$x=\pm \frac{\sqrt{33}}{-8}$

x=4,x=-9

$x=9+\frac{\sqrt{33}}{8},x = 9-\frac{\sqrt{33}}{8}$

$x= \frac{4\pm \sqrt{65}}{2}$

$x =\frac{9\pm \sqrt{33}}{8}$

Correct answer:

$x =\frac{9\pm \sqrt{33}}{8}$

Explanation:

4x^2-9x+3

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

where a, b, and c are the constants defined as follows:

ax^2+bx+c = 0

This means our a is 1, our b is -6, and our c is 8.

Finally, lets plug the numbers into the formula:

$\frac{-(-9)\pm \sqrt{(-9)^2-4(4)(3)}}{2(4)}$

$\frac{9\pm \sqrt{81-48}}{8}$

$\frac{9\pm \sqrt{33}}{8}$

These are the roots of the equation.

$x =\frac{9\pm \sqrt{33}}{8}$

Remember that using $\pm$ is the exact same as writing:

$x =\frac{9+ \sqrt{33}}{8}, x =\frac{9- \sqrt{33}}{8}$

Report an Error

Example Question #491 : Intermediate Single Variable Algebra

Use the quadratic formula to find the answer of the following quadratic equation.

3x^2-9x+2

Possible Answers:

$x=3+\sqrt{57},x=3-\sqrt{57}$

$x=\frac{3+\sqrt{57}}{2},x=\frac{3-\sqrt{57}}{2}$

$x=\frac{9+\sqrt{57}}{6},x=\frac{9-\sqrt{57}}{6}$

$x=\frac{9+2\sqrt{57}}{6},x=\frac{9-2\sqrt{57}}{6}$

$x=\frac{9+\sqrt{59}}{6},x=\frac{9-\sqrt{59}}{6}$

Correct answer:

$x=\frac{9+\sqrt{57}}{6},x=\frac{9-\sqrt{57}}{6}$

Explanation:

The quadratic equation is:

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

$\\a=3 \\b=-9 \\c=2$

Therefore:

$\frac{-(-9)\pm \sqrt{(-9)^2-4(3)(2)}}{2(3)}$

Which gives the answer:

$x=\frac{9\pm \sqrt{57}}{6}$

Report an Error

Example Question #31 : Quadratic Formula

Solve for x:

3x^2+12x+15=0

Possible Answers:

$x=-2\pm i$

x=-2+i

$x=2\pm i$

$x=-12\pm 6i$

Correct answer:

$x=-2\pm i$

Explanation:

For a quadratic function

ax^2+bx+c

the quadratic formula states that

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Using the formula for our function, we get

$x=\frac{-12\pm \sqrt{(12)^2-4(3)(15)}}{2(3)}=\frac{-12\pm \sqrt{-36}}{6}$

Notice that we have a negative under the square root. This means that we must use the imaginary number

$i=\sqrt{-1}$

and our roots will be imaginary.

Simplifying using i, we get

$x=\frac{-12\pm 6i}{6}=-2\pm i$

Report an Error

Example Question #179 : Solving Quadratic Equations

Find the roots using the quadratic formula

x^2+7x+7

Possible Answers:

$-7 \pm \sqrt{21}$

$\frac{-7\pm \sqrt{21}}{2}$

$\frac{7\pm \sqrt{21}}{2}$

$\frac{-7}{2} \pm \sqrt{21}$

$7 \pm \sqrt{21}$

Correct answer:

$\frac{-7\pm \sqrt{21}}{2}$

Explanation:

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

For this problem

a=1, the coefficient on the x^2 term

b=7, the coefficient on the x term

c=7, the constant term

$\frac{-7\pm \sqrt{7^2-4(1)(7)}}{2(1)}\Rightarrow \frac{-7\pm \sqrt{21}}{2}$

Report an Error

Example Question #180 : Solving Quadratic Equations

Find a root for: y=15x^2+1

Possible Answers:

$i\sqrt{15}$

$\textup{There are no possible roots.}$

$-\frac{i\sqrt{15}}{15}$

$2i\sqrt{15}$

-5i

Correct answer:

$-\frac{i\sqrt{15}}{15}$

Explanation:

Write the quadratic equation that applies for y=ax^2+bx+c .

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

There is no term. Substitute the known coefficients from the polynomial.

$x=\frac{\pm \sqrt{-4(15)(1)}}{2(15)}$

Simplify the numerator and denominator.

$x=\frac{ \pm\sqrt{-60}}{30} = \pm\frac{\sqrt{-1} \cdot 2 \sqrt{15}}{30} = \pm \frac{i\sqrt{15}}{15}$

The roots will be imaginary.

One of the possible answers is: $-\frac{i\sqrt{15}}{15}$

Report an Error

View Algebra 2 Tutors

John
Certified Tutor

Arizona State University, Bachelor of Science, Microbiology.

View Algebra 2 Tutors

Lawrence
Certified Tutor

Harvey Mudd College, Bachelor of Engineering, Engineering, General. UC Berkeley, Master's/Graduate, Engineering.

View Algebra 2 Tutors

Darryl
Certified Tutor

University of Florida, Bachelor of Science, Mathematics.

All Algebra II Resources

10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in Boston, Reading Tutors in Dallas Fort Worth, SSAT Tutors in New York City, Computer Science Tutors in San Francisco-Bay Area, GMAT Tutors in Washington DC, GRE Tutors in Los Angeles, English Tutors in Seattle, SAT Tutors in Houston, Statistics Tutors in Atlanta, Calculus Tutors in Phoenix

Popular Courses & Classes

GMAT Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in Miami, MCAT Courses & Classes in Houston, ISEE Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Seattle, ACT Courses & Classes in Chicago, Spanish Courses & Classes in Boston, Spanish Courses & Classes in Phoenix, SSAT Courses & Classes in Denver, GRE Courses & Classes in Chicago

Popular Test Prep

LSAT Test Prep in Miami, LSAT Test Prep in Seattle, GMAT Test Prep in San Diego, MCAT Test Prep in Seattle, MCAT Test Prep in Chicago, GRE Test Prep in Denver, SAT Test Prep in Phoenix, ISEE Test Prep in Chicago, LSAT Test Prep in Atlanta, SSAT Test Prep in Philadelphia

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

Algebra II : Solving Quadratic Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #31 : Quadratic Formula

Example Question #172 : Solving Quadratic Equations

Example Question #1621 : Algebra Ii

Example Question #174 : Solving Quadratic Equations

Example Question #31 : Quadratic Formula

Example Question #171 : Solving Quadratic Equations

Example Question #491 : Intermediate Single Variable Algebra

Example Question #31 : Quadratic Formula

Example Question #179 : Solving Quadratic Equations

Example Question #180 : Solving Quadratic Equations

All Algebra II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: