Welcome to our community

Be a part of something great, join today!

Understanding how to deal with fractions using brackets

Status
Not open for further replies.

Casio

Member
Feb 11, 2012
86
(- 5x/3 + 2/3) (- 5x/3 + 2/3)

If the above example was;

(-5x + 2) (-5x + 2) = 25x^2 - 10x - 10x + 4 =

25x^2 - 20x + 4

The problem is I don't know how to deal with the denominators in this form?

Anyone help
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
(- 5x/3 + 2/3) (- 5x/3 + 2/3)

If the above example was;

(-5x + 2) (-5x + 2) = 25x^2 - 10x - 10x + 4 =

25x^2 - 20x + 4

The problem is I don't know how to deal with the denominators in this form?

Anyone help
You can think of it this way. Let $a=-\frac{5}{3}$ and $b=\frac{2}{3}$.

Then

\[\left(-\frac{5}{3}x+\frac{2}{3}\right)\left(-\frac{5}{3}x+\frac{2}{3}\right)=(ax+b)(ax+b)=a^2x^2+2abx+b^2\]

Now substitute in the values for a and b to get the answer.

Another way to do this would be to notice that

\[\left(-\frac{5}{3}x+\frac{2}{3}\right)\left(-\frac{5}{3}x+\frac{2}{3}\right)=\frac{1}{3}(-5x+2)\cdot\frac{1}{3}(-5x+2)=\frac{1}{9}(-5x+2)(-5x+2)\]

Thus, the answer would be what you found for $(-5x+2)(-5x+2)$ multiplied by $\frac{1}{9}$, i.e.

\[\frac{1}{9}\left(25x^2 - 20x + 4\right) = \frac{25}{9}x^2-\frac{20}{9}x+\frac{4}{9}\]

I hope this helps!
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
(- 5x/3 + 2/3) (- 5x/3 + 2/3)

If the above example was;

(-5x + 2) (-5x + 2) = 25x^2 - 10x - 10x + 4 =

25x^2 - 20x + 4

The problem is I don't know how to deal with the denominators in this form?

Anyone help
The shortcuts Chris L T521 mentioned are awesome but if you don't see how they work then this is the much slower way but the method always works with these problems.

You FOIL the same way you do for the second example. So the problem for you lies in how to multiply fractions I think. Luckily, multiplying fractions is the easiest operation to remember, unlike adding them. You just multiple the top line together and that's the numerator of the new fraction and you multiply the bottom line together and that's the denominator of the new fraction. For example:

\(\displaystyle \frac{1}{2} \times \frac{2}{3}=\frac{2}{6}=\frac{1}{3}\)

The two top parts are multiplied together and the two bottom parts are as well. Then the last step is just simplification.

If you have a variable, $x$, in the fraction as well then you need to make sure you only multiply like things when calculating your result. For example:

\(\displaystyle \frac{3x}{2} \times \frac{6}{5}=\frac{18x}{10}=\frac{9x}{5}\)

The numbers multiply with numbers only and the variable, $x$ comes along for the ride. If there is more than one $x$ then you need to do another step. For example:

\(\displaystyle \frac{2x}{5} \times \frac{3x}{2}=\frac{6x^2}{10}=\frac{3x^2}{5}\)

You have 3 ways to get the answer now so try one out :)
 
Last edited:

Casio

Member
Feb 11, 2012
86
You can think of it this way. Let $a=-\frac{5}{3}$ and $b=\frac{2}{3}$.

Then

\[\left(-\frac{5}{3}x+\frac{2}{3}\right)\left(-\frac{5}{3}x+\frac{2}{3}\right)=(ax+b)(ax+b)=a^2x^2+2abx+b^2\]

Now substitute in the values for a and b to get the answer.

Another way to do this would be to notice that

\[\left(-\frac{5}{3}x+\frac{2}{3}\right)\left(-\frac{5}{3}x+\frac{2}{3}\right)=\frac{1}{3}(-5x+2)\cdot\frac{1}{3}(-5x+2)=\frac{1}{9}(-5x+2)(-5x+2)\]

Thus, the answer would be what you found for $(-5x+2)(-5x+2)$ multiplied by $\frac{1}{9}$, i.e.

\[\frac{1}{9}\left(25x^2 - 20x + 4\right) = \frac{25}{9}x^2-\frac{20}{9}x+\frac{4}{9}\]

I hope this helps!
I have followed through all the above and can see what you have done with the denominator 3, where you have ended up with my original solution, but I didn't know what to do with the denominators.

If then \[\frac{1}{9}\left(25x^2 - 20x + 4\right)\] is equal to what I had originally, from this point should I multiply everything inside the brackets by \[\frac{1}{9}\] ?

If I were to subtract 20x from 25x^2 this would leave 5x + 4, would this be the correct way?

So how I have to add x^2 + 6x + 9 to the above and I think that loosing the fractions is the correct method to use, therefore I can then add all like terms to get a solution.
 
Last edited by a moderator:

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
If I were to subtract 20x from 25x^2 this would leave 5x + 4, would this be the correct way?
You can't combine $x$ and $x^2$ by addition or subtraction. The solution will have at least a term with $x^2$, a term with $x$ and a constant. If you already factored out the \(\displaystyle \frac{1}{9}\) then I think it's ok to call that the final answer but you can also distribute it to the other terms and get rid of the parentheses. That's the only thing left you can do to simplify the expression, so you are more or less done :)
 

Casio

Member
Feb 11, 2012
86
You can't combine $x$ and $x^2$ by addition or subtraction. The solution will have at least a term with $x^2$, a term with $x$ and a constant. If you already factored out the \(\displaystyle \frac{1}{9}\) then I think it's ok to call that the final answer but you can also distribute it to the other terms and get rid of the parentheses. That's the only thing left you can do to simplify the expression, so you are more or less done :)
I can't see that being the end of the problem?

X^2 + 6x + 9 + (1\9)(25X^2 - 20x + 4)

Something is not quite finished as I require to find the coordinates of any points of my circle to finish, therefore the fraction (1\9) must go so that I can add and subtract terms to end up with an expression that can be either factored or use the formula to find values for x, thus I can then plug them into 3y = -5x + 14 to determine the value of y, and then finally decide whether the line intersects the circle and if so show the roots.

The RHS is still confusing to me?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
What is the full problem you are trying to solve? I can't make sense of your post until I know that. My previous post was talking about using the FOIL method to remove the parentheses on \(\displaystyle \left(-\frac{5}{3}x+\frac{2}{3}\right)\left(-\frac{5}{3}x+\frac{2}{3}\right)\) and then simplifying the result as much as possible.

What is supposed to be equal to what? :confused: Please post the whole question and we'll get to the bottom of it. :)

The main point of my previous post was commenting that \(\displaystyle (25x^2-20x) \ne 5x\) for any $x$ and so you can't simplify the expression that way.
 
Last edited:

Casio

Member
Feb 11, 2012
86
What is the full problem you are trying to solve? I can't make sense of your post until I know that. My previous post was talking about using the FOIL method to remove the parentheses on \(\displaystyle \left(-\frac{5}{3}x+\frac{2}{3}\right)\left(-\frac{5}{3}x+\frac{2}{3}\right)\) and then simplifying the result as much as possible.

What is supposed to be equal to what? :confused: Please post the whole question and we'll get to the bottom of it. :)

The main point of my previous post was commenting that \(\displaystyle (25x^2-20x) \ne 5x\) for any $x$ and so you can't simplify the expression that way.
OK from this point onwards some people could now say that this is going to be in the wrong thread if I post as this is a circle problem, so I guess I should ask if I am permitted to continue or start another thread in geometry?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
OK from this point onwards some people could now say that this is going to be in the wrong thread if I post as this is a circle problem, so I guess I should ask if I am permitted to continue or start another thread in geometry?
Thank you for being so aware of our forum rules! You make a good point :) I think you can start another thread for your question and I'll close this one now. However if you had made a post with the circle problem you wouldn't get an infraction since I specifically asked you to do so. :)
 
Status
Not open for further replies.