MHB Find the product of all real solutions

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The discussion revolves around finding the product of all real solutions to the equation involving multiple square roots. Participants confirm that they have arrived at the correct answer, indicating successful collaboration. One member shares their recovery from food poisoning, adding a personal touch to the conversation. The focus remains primarily on solving the mathematical equation. Overall, the thread highlights both problem-solving and community engagement.
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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
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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