Making x the subject in two equations

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  • Thread starter Thread starter Ellie1
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Discussion Overview

The discussion revolves around the process of making the variable x the subject in two equations: \(y=2x^2 +3\) and \(y=\sqrt{\frac{x}{3}}\). The scope includes mathematical reasoning and problem-solving related to algebraic manipulation.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks how to isolate x in the equations provided.
  • Another participant demonstrates the steps to isolate x in the first equation, concluding that \(x = \pm \sqrt{\frac{y-3}{2}}\) and notes the condition \(y \geq 3\) for the solution to be valid.
  • A third participant suggests that it is important to show what has been attempted when asking for help, indicating a preference for collaborative problem-solving.
  • One participant reiterates the solution for the first equation and emphasizes the condition \(y \geq 3\), agreeing that it is relevant to the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the steps to isolate x in the first equation and the importance of the condition \(y \geq 3\). However, the second equation remains unaddressed, and no consensus is reached regarding its manipulation.

Contextual Notes

The second equation \(y=\sqrt{\frac{x}{3}}\) has not been explored in detail, and there are no provided steps or conditions related to it.

Ellie1
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how do you make x the subject in
$$y=2x^2 +3$$

how do you make x the subject in
$$y=\sqrt{x \over 3}$$
 
Last edited by a moderator:
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Ellie said:
how do u make x the subject in
y=2x squared +3
(Wave)

$y= 2x^2+3 \Rightarrow 2x^2=y-3 \Rightarrow x^2=\frac{y-3}{2} \Rightarrow x= \pm \sqrt{\frac{y-3}{2}}$

Notice that it has to be $y-3 \geq 0 \Rightarrow y \geq 3$.
Ellie said:
how do you make x the subject in
y=\sqrt{xover3

Give it a try? (Thinking)
 
Last edited:
Hello, Ellie and welcome to MHB! :D

I have moved your thread since our "Introductions" subforum is meant for folks to post a bit about themselves to let our community know a bit about you.

I have retitled your thread so that is briefly describes the nature of the questions being asked.

We also ask that when you post questions, you show what you have tried so our helpers can see where you are stuck or what you may be doing wrong, and this way we can offer better help.

I am thinking your first equation is:

$$y=2x^2+3$$

What do you think is the first thing we should do in an effort to solve for $x$?
 
evinda said:
(Wave)

$y= 2x^2+3 \Rightarrow 2x^2=y-3 \Rightarrow x^2=\frac{y-3}{2} \Rightarrow x= \pm \sqrt{\frac{y-3}{2}}$

Notice that it has to be $y-3 \geq 0 \Rightarrow y \geq 3$.
Perhaps you had already noticed this, and I would agree that it is good to point out [math]y \geq 3[/math] anyway, but I should point out that this condition is satisfied by the given equation.

-Dan
 

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