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- Feb 14, 2012

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

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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

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

- Nov 4, 2013

- 428

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

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 $\boxed{\dfrac{81}{2}}$

- Nov 29, 2013

- 172

Hello.

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

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

[tex]\forall{y}>40.5 \ and \ \forall{y}<40.5 \ \rightarrow{}[/tex]

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

[tex]key \ term=\sqrt{81 \pm{2y}}[/tex]

Regards

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- Feb 14, 2012

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Yes, I have recovered fully from food poisoning and thanks for asking!Hello.

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