Solving algebraic fractions using quadratics

[Simonsky]

Here's a the question:

x^2 +3 /4 +(2x-1)/5 = 1 . For some reason the first numerator (x^2 + 3) is not in brackets, not sure there is any significance in that.

Here is what I tried:

1. I cross multiplied: 5(x^2 + 3) + 4(2x -1) =1 = 5x^2 +15 + 8x - 4 =1 = 5x^2 +8x +10 = 0

I can already see that if I use the completing the square formula I will end up with a negative discriminant: sqrt: 64 -200.

I'm obviously doing something wrong!

 

[Dr.Peterson]

Here's a the question:

x^2 +3 /4 +(2x-1)/5 = 1 . For some reason the first numerator (x^2 + 3) is not in brackets, not sure there is any significance in that.

Here is what I tried:

1. I cross multiplied: 5(x^2 + 3) + 4(2x -1) =1 = 5x^2 +15 + 8x - 4 =1 = 5x^2 +8x +10 = 0

I can already see that if I use the completing the square formula I will end up with a negative discriminant: sqrt: 64 -200.

I'm obviously doing something wrong!

If x^2 + 3 is not in brackets, and the expression is written in-line as you did, then it is not a numerator! What you wrote means \(\displaystyle x^2 + \frac{3}{4} + \frac{2x-1}{5}=1\).

If it was really \(\displaystyle \frac{x^2 + 3}{4} + \frac{2x-1}{5}=1\), where no parentheses are needed because the fraction bar serves the purpose, then your "cross-multiplication" is wrong; in effect you have multiplied the left side by 20 but did not multiply the right side by 20 as well, so you have changed the equation.

I avoid talking about cross-multiplication, because too many students think they are doing it when it really doesn't apply, as here. You can think of addition of fractions as cross-multiplication if you put the product of the denominators as the new denominator. You are probably think of the the cross-multiple that applies when you are solving a proportion, which is an entirely different situation. It's far better just to always think in terms of multiplying both sides by the same quantity.

Also, the equal signs I put in red above are wrong. Don't use "=" to mean "therefore"; it only means that two quantities are equal, not that two equations are related. Some people use an arrow, like "=>" for this purpose (meaning "implies"), which can be appropriate.

 

[Denis]

x^2 +3 /4 +(2x-1)/5 = 1 .
For some reason the first numerator (x^2 + 3) is not in brackets

x^2 + 3 is NOT a numerator.
Equation really is:
x^2 + (2x-1)/5 = 1 - 3/4
SO:
x^2 + (2x-1)/5 = 1/4

Continue...hint: multiply equation by 20

 

[Simonsky]

If x^2 + 3 is not in brackets, and the expression is written in-line as you did, then it is not a numerator! What you wrote means \(\displaystyle x^2 + \frac{3}{4} + \frac{2x-1}{5}=1\).

If it was really \(\displaystyle \frac{x^2 + 3}{4} + \frac{2x-1}{5}=1\), where no parentheses are needed because the fraction bar serves the purpose, then your "cross-multiplication" is wrong; in effect you have multiplied the left side by 20 but did not multiply the right side by 20 as well, so you have changed the equation.

I avoid talking about cross-multiplication, because too many students think they are doing it when it really doesn't apply, as here. You can think of addition of fractions as cross-multiplication if you put the product of the denominators as the new denominator. You are probably think of the the cross-multiple that applies when you are solving a proportion, which is an entirely different situation. It's far better just to always think in terms of multiplying both sides by the same quantity.

Also, the equal signs I put in red above are wrong. Don't use "=" to mean "therefore"; it only means that two quantities are equal, not that two equations are related. Some people use an arrow, like "=>" for this purpose (meaning "implies"), which can be appropriate.

Yes, thanks and sorry for the typographical errors-I really need to make the effort to learn LATEX. I forgot to add the top parts over the same denominator. so, I think, I was cross-multiplying at too early a stage:

With a denominator of 20 it becomes: (5x^2 + 8x + 11)/20 = 1 which can then be cross multiplied to get 5x^2 + 8x -9 = 0.

I forgot the decisive stage .

 

[mmm4444bot]

… sorry for the typographical errors -- I really need to make the effort to learn LATEX …

Here's a note about symbolism because I don't want other readers to be confused; LaTex is not required to properly communicate math expressions. With basic expressions (like yours), it's much easier to simply use proper grouping symbols.

:idea: There is a link in the forum guidelines to a web site that explains how to type math as text.

PS: I changed your single hyphen to a dash (shown in red). Don't type a single hyphen when you want a dash. Two hyphens comprise a dash. In this forum, a single hyphen means: a subtraction operator, a negation symbol, or a direction indicator. :cool: