(Solved) Solve this quadratic equation(X^2+14x+33=0)?

darkyoda1589

wrote...

Go to Answer

Posts: 4

Rep:

11 years ago

Solve this quadratic equation(X^2+14x+33=0)?

X^2+14x+33=0

I solved it as ss={11,3}
but on mathway (http://mathway.com/problem.aspx?p=precalculus) it says {-11,-3}

which one is right?
Oh,I know know what went wrong.
I solved by quadratic equation, and I began it with a b,I simply forgot the (-).

:")
Thanks all.

Read 465 times

5 Replies

Report

Replies

robbaron

wrote...

11 years ago

x^2+14x+33
(x+11)(x+3)=0
x+11=0; x+3=0
x=-11; x=-3

The latter.

Report

Answer accepted by topic starter

deep3636

wrote...

Posts: 4

Rep:

11 years ago

Report

Related Topics

rmalviya

wrote...

11 years ago

I think where you went wrong is when you had it (correctly) factored as
(x+11)(x+3) = 0
..you forgot to change the signs...
If we multiply 2 numbers and get 0, then at least one of those numbers must be 0, so
x+11 = 0 or x+3 = 0
Solving these for x, we get
x = -11 or -3

Report

socalkid1

wrote...

11 years ago

x^2+14x+33=0

Factor the "c" term which is 33.

We get: 1*33 and 3*11

We will use 3*11 because that is the factor pair that adds up to the "b" term, which is 14. (3+11=14)

So this equation factors into:
(x+3)(x+11)=0

Using the zero product rule, we can split this into two equations:
x+3=0 and x+11=0

Solving for x in both equations, you get the two solutions to be:
x=-3 and x=-11

Report

robbcfd

wrote...

11 years ago

One hint that may help you to check your work in the future is to look at the signs in the equation. Since the second sign is positive, you know both signs in the factored equation must be the same. Add to this information that the first sign is positive and you know that when you factor the equation, both signs will be positive (and thus both the solutions must be negative). If the first sign of the original equation had been negative and the second positive, both signs in the factored equation would be negative and the solutions both positive. However if the second sign in the original equation were negative, the factored solution would have one positive and one negative sign with a positive and negative solution. This would have shown you that there was a problem with your ss={11,3} . The information can be very helpful when taking timed tests. You can quickly eliminate wrong answers and may not need to spend time working the whole solution.

Report

New Topic

Home Search Q & A Board Gallery Resource Library Blog Dictionary Chat	Quick Links Start a Topic Latest Topics Unanswered Questions Courses Study Tools Browse by Textbook	Social Facebook YouTube Community Stats Who's Online Trending Staff	Help About Us Take the Tour Study Tips Posting Guidelines Terms and Policies Contact Us
Biology Forums - Study Force © 2010-2024 \| Powered by SMF \| SMF © 2015, Simple Machines \| Sitemap Biology-Forums.com is not affiliated with any publisher. Book covers, title and author names appear for reference only.