How to rationalize the numberator?

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Discussion Overview

The discussion revolves around the process of rationalizing the numerator of expressions involving square roots, specifically focusing on the fractions \(\frac{\sqrt{x} - 3}{x - 9}\) and \(\frac{\sqrt{x} - 2}{4 - x}\). Participants seek clarification and assistance in understanding how to perform this operation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants express confusion about the correct interpretation of the original questions regarding the expressions.
  • There is a proposal to use the identity \(a^2 - b^2 \equiv (a+b)(a-b)\) to facilitate the rationalization process.
  • One participant suggests multiplying both the numerator and denominator by the conjugate of the numerator to eliminate the square root from the numerator.
  • Another participant provides a specific algebraic manipulation involving the expressions but questions whether their approach is correct.
  • There is a mention of merging threads due to misclassification of the topic within the forum.

Areas of Agreement / Disagreement

Participants generally agree on the expressions that need to be rationalized, but there is no consensus on the specific methods or steps to achieve this. Confusion and differing interpretations of the original questions persist.

Contextual Notes

Some participants may be missing foundational concepts such as the definition and use of conjugates in rationalization, which could affect their understanding of the problem.

dtcool2003
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Here is the question: the square root of X minus 3, but the 3 is not under the square root divided by x-9? also another question: the square root of X minus 2, but the 2 is not under the square root divided by 4-x? anyone can help me?
 
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How to rationalize the numerator?

Here is the question: sqrt(x) - 3/ x-9 ? also another question: sqrt(x) -2/ 4-x ? anyone can help me? Thanks
 
Last edited:
So you mean sqrt(x)/(x-9) - 3/(x-9)?
 
No

no I mean, sqrt(x) - 3/ x-9 ? also another question: sqrt(x) -2/ 4-x ? anyone can help me? Thanks and I need to rationalize the numerator
 
Nope, i guess he means

[tex]\frac{\sqrt{x} -3}{x-9}[/tex] and [tex]\frac{\sqrt{x} -2}{4-x}[/tex]

When rationalizing, remember that [itex]a^2 - b^2 \equiv (a+b)(a-b)[/itex]. So take [/itex] a=\sqrt{x} [/itex]. What is 'b' equal to ?
 
yeah

those are the right equations u just put up, but i don't know how to rationalize those equations?
 
yeah

I got x-9/ x-9sqrt(x) + 3x - 27 is that right?
 
[tex]\frac{\sqrt{x}-3}{x^2- 9}[/tex]
That it?

Use (a- b)(a+ b)= a2- b2: multiply both numerator and denominator by [itex]\sqrt{3}+ 3[/itex] and the squareroot will disappear from the numerator. It will, of course, show up in the denominator.

For the second, assuming you have a fraction with [itex]\sqrt{x}- 2[/itex] in either numerator or denominator, multiply both numerator and denominator by [itex]\sqrt{x}+ 2[/itex]
 
This is not a "Linear and Abstract Algebra" question and it was also posted in the "General Math" section so I am merging the two threads in that section.
 
  • #10
[tex]\frac {\sqrt{x}-3}{x-9}[/tex]

Multiply:
Both [tex]{x-9}[/tex] and [tex]{\sqrt{x}-3}[/tex]

By the conjugate of [tex]{\sqrt{x}-3}[/tex] which is [tex]{\sqrt{x}+3}[/tex].

[tex]\frac {\sqrt{x}-3( \sqrt{x}+3)} {x-9( \sqrt{x}+3)}[/tex]

From here, you should be able to do some algebra, to rationalize the numerator. Do you know what a conjugate is, and why it is used? That might be your issue with not understanding it.
 

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