Find the product of all real solutions

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Discussion Overview

The discussion revolves around finding the product of all real solutions to a specific equation involving square roots and polynomial expressions. The focus is on the mathematical reasoning and problem-solving aspects related to this equation.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Participants present the same equation multiple times, indicating a focus on solving it.
  • One participant expresses gratitude for contributions and suggests that the correct answer has been reached.
  • Casual conversation occurs among participants, including inquiries about health, which may indicate a friendly community atmosphere.

Areas of Agreement / Disagreement

There appears to be some agreement that the correct answer has been identified, but the specifics of the solution and the methods used remain unclear. The discussion does not resolve the mathematical details of the problem.

Contextual Notes

The repeated posting of the same equation may suggest a lack of clarity or confusion regarding the problem, but no specific assumptions or limitations are explicitly stated.

Who May Find This Useful

Participants interested in solving complex equations involving square roots and polynomials, as well as those looking for a collaborative problem-solving environment.

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Find the product of all real solutions to the equation

$\sqrt{2y-81}-\sqrt{1734-20y+4y^2}-\sqrt{81-2y}+\sqrt{4y^2+26y-129}=0$.
 
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anemone said:
Find the product of all real solutions to the equation

$\sqrt{2y-81}-\sqrt{1734-20y+4y^2}-\sqrt{81-2y}+\sqrt{4y^2+26y-129}=0$.

From the first surd, we have the condition $y \geqslant \frac{81}{2}$.

From the third surd, we have the condition $y \leqslant \frac{81}{2}$. But from the first condition, we have only one possible value of y for which both these expressions are defined i.e $\frac{81}{2}$

Substituting 81/2 in the given equation makes the first and the third surd equal to zero. Now we have to check if the remaining two surds give the same value when y=81/2 is substituted. Directly substituting 81/2 takes time (at least for me) so we take the reverse path. We equate the expressions inside the remaining surds to see at what value of y they are equal, if y comes out to be 81/2, then 81/2 is the answer.

$$4y^2+26y-129=4y^2-20y+1734 \Rightarrow 46y=1863 \Rightarrow y=\frac{81}{2}$$

Hence, the answer is $\boxed{\dfrac{81}{2}}$
 
anemone said:
Find the product of all real solutions to the equation

$\sqrt{2y-81}-\sqrt{1734-20y+4y^2}-\sqrt{81-2y}+\sqrt{4y^2+26y-129}=0$.

Hello.

Has you healed?. I'd like to that you have recovered.

y=40.5

\forall{y}>40.5 \ and \ \forall{y}<40.5 \ \rightarrow{}

\rightarrow{} \sqrt{2y-81}-\sqrt{1734-20y+4y^2}-\sqrt{81-2y}+\sqrt{4y^2+26y-129} \neq{0}

key \ term=\sqrt{81 \pm{2y}}

Regards
 
Thanks to both of you for participating and you have gotten the right answer! Well done! :)
mente oscura said:
Hello.

Has you healed?. I'd like to that you have recovered.

Yes, I have recovered fully from food poisoning and thanks for asking!:o
 

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