Welcome to our community

Be a part of something great, join today!

Help Please :) Quadratic equation

babcockkw

New member
Mar 28, 2013
4
So here is the question:
2x/7=2x^2
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Help Please :)

When our members post questions, we expect for them to show what they have tried, or what they think they should do, or at the very least state that they simply have no idea how to begin. This way our helpers no where the person is stuck and can offer guidance at that point.

Can you show what you have tried and where you are stuck?

edit: I have edited the topic title to add a bit of information regarding the nature of the question you are asking. A title stating that you are asking for help provides no information, as the fact that you are posting implies that you are seeking help. The title should include more information than can be discerned from the forum in which you are posting. For example the title Please help with this trig. problem, posted in the Trigonometry forum does not give our members any information, but a title such as How do I solve this problem using the Law of Cosines tells people something about the problem within the topic. This benefits our members in that simply by browsing the forums they can see what kinds of topics are addressed, and searches are also made easier with such descriptive titles. :D
 
Last edited:

babcockkw

New member
Mar 28, 2013
4
Re: Help Please :)

When our members post questions, we expect for them to show what they have tried, or what they think they should do, or at the very least state that they simply have no idea how to begin. This way our helpers no where the person is stuck and can offer guidance at that point.

Can you show what you have tried and where you are stuck?
Absolutely, I have tried to break up the right side of the equation, so that the equation looks like: 2x/7= (2x)(2x)
After this, I divided each side by 2x:
(2x/7)/2x=(2x^2)/2x
After that, I'm pretty stuck.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The first thing I would do is observe that both sides have 2 as a factor, so I would divide through by 2. What does your equation look like now?
 

babcockkw

New member
Mar 28, 2013
4
The first thing I would do is observe that both sides have 2 as a factor, so I would divide through by 2. What does your equation look like now?
Starting from the beginning or where I currently have the equation?

Alright, maybe I figured it out. I got x=7. Let me explain:
2x/7=2x^2 Divide each side by two
(2x/7)/2=(2x^2)/2 simplify to:
x/7=x^2 Divide each side by x
(x/7)/x= (x^2)/x Simplify to:
7=x
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I'm sorry, from the beginning:

\(\displaystyle \frac{2x}{7}=2x^2\)
 

babcockkw

New member
Mar 28, 2013
4
I'm sorry, from the beginning:

\(\displaystyle \frac{2x}{7}=2x^2\)
Alright, maybe I figured it out. I got x=7. Let me explain:
2x/7=2x^2
Divide each side by two
(2x/7)/2=(2x^2)/2
simplify to:
x/7=x^2
Divide each side by x
(x/7)/x= (x^2)/x
Simplify to:
7=x

I'm a bit confused on when you can cancel while dividing. Is it just when the top & bottom mach because they would just equal one?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Well, $x=7$ does not work...try substituting it into the original equation and you will find:

\(\displaystyle \frac{2(7)}{7}=2(7)^2\)

\(\displaystyle 2=98\)

This is not true, so that solution must be incorrect. Also since the equation is quadratic, you should expect to find 2 solutions. When you divide through by $x$ you need to be aware that you are eliminating $x=0$ as a solution. I don't like to tell students to divide through by $x$ at this level of math, although opinions on this will vary.

So, you did correctly divide through by 2 to obtain:

\(\displaystyle \frac{x}{7}=x^2\)

Next, I suggest multiplying through by 7 to get rid of that denominator...what do you get?
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Alright, maybe I figured it out. I got x=7. Let me explain:
2x/7=2x^2
Divide each side by two
(2x/7)/2=(2x^2)/2
simplify to:
x/7=x^2
Divide each side by x
(x/7)/x= (x^2)/x
Simplify to: (mistake going from above to below)
7=x

I'm a bit confused on when you can cancel while dividing. Is it just when the top & bottom mach because they would just equal one?
[JUSTIFY]And for the record, this approach is essentially correct (with MarkFL's warning that dividing by zero implies that x is not zero, so you need to take that into account and check whether zero is in fact a solution, this often confuses students and so is not generally taught until later) but you made a mistake in your computation (in red), leading you to a wrong solution. Can you see the error?[/JUSTIFY]
 

Petrus

Well-known member
Feb 21, 2013
739
Hello,
This how I would solve it:) (I Will explain with words) When you divide by x you can think Two case cause if you type 3/0 in your calculator it Will give you 'undefined' so we can think like Two case. (Notice you got x^2 that means you Will get more Then one soloution)
case 1:
when x=0. That one is pretty simple, just replace x with 0 and look if it's true. In your equation you Will get 0=0 and that is true, that means we got à soloution when x is 0!:)
Case 2
when x is NOT equal to zero. That means you can divide by x:) post when you got soloution or need help or got any question:)
 
Last edited: