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605-655 Level|   Roots|      
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Bunuel
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The expression (1/((8 - 63^(1/2))^(1/2) + 1/((8 + 63^(1/2))^(1/2))^2 = 16 Nov 2021, 09:45
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45% (medium)

Question Stats:

69% (02:19) correct 31% (02:37) wrong based on 149 sessions

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The expression \((\frac{1}{\sqrt{8 - \sqrt{63}}} +\frac{1}{\sqrt{8 +\sqrt{63}}})^2 =\)

(A) 21
(B) 20
(C) 19
(D) 18
(E) 17
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The expression (1/((8 - 63^(1/2))^(1/2) + 1/((8 + 63^(1/2))^(1/2))^2 = 16 Nov 2021, 10:08
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The expression \((\frac{1}{\sqrt{8 - \sqrt{63}}} +\frac{1}{\sqrt{8 +\sqrt{63}}})^2 =\)

(A) 21
(B) 20
(C) 19
(D) 18
(E) 17


Multiply by their conjugates to remove denominator.


\((\frac{1}{\sqrt{8 - \sqrt{63}}} +\frac{1}{\sqrt{8 +\sqrt{63}}})^2 =\)

\((\frac{ \sqrt{8 + \sqrt{63}}}{\sqrt{8 - \sqrt{63}}* \sqrt{8 + \sqrt{63}}} +\frac{ \sqrt{8 - \sqrt{63}}}{\sqrt{8 +\sqrt{63}}* \sqrt{8 - \sqrt{63}}})^2 =\)

\((\frac{ \sqrt{8 + \sqrt{63}}}{\sqrt{64 - {63}}} +\frac{ \sqrt{8 - \sqrt{63}}}{\sqrt{64 -{63}}})^2 =\)

\( (\sqrt{8 + \sqrt{63}}+\sqrt{8 - \sqrt{63}})^2\)

\( (\sqrt{8 + \sqrt{63}})^2+(\sqrt{8 - \sqrt{63}})^2\) \( +2(\sqrt{8 + \sqrt{63}}*\sqrt{8 - \sqrt{63}})\)

\(8+\sqrt{63}+8-\sqrt{63}+2=18\)

OR

\((\frac{1}{\sqrt{8 - \sqrt{63}}} +\frac{1}{\sqrt{8 +\sqrt{63}}})^2 =\)

\((\frac{ \sqrt{8 +\sqrt{63}}+ \sqrt{8 - \sqrt{63}}}{\sqrt{8 - \sqrt{63}}*\sqrt{8 +\sqrt{63}}})^2 =\)

\( (\sqrt{8 + \sqrt{63}}+\sqrt{8 - \sqrt{63}})^2\)

\( (\sqrt{8 + \sqrt{63}})^2+(\sqrt{8 - \sqrt{63}})^2\) \(+2 (\sqrt{8 + \sqrt{63}}*\sqrt{8 - \sqrt{63}})\)

\(8+\sqrt{63}+8-\sqrt{63}+2=18\)
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The expression (1/((8 - 63^(1/2))^(1/2) + 1/((8 + 63^(1/2))^(1/2))^2 = Updated on: 17 Feb 2024, 05:50
Originally posted by ruis on 17 Feb 2024, 04:59.
Last edited by ruis on 17 Feb 2024, 05:50, edited 3 times in total.
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Hi chetan2u, thank you for sharing your approach! I tried solving using identities as a first approach.
I then tried solving with your approach (multiplying by conjugate) and I am still getting it wrong. Can´t see where I commited the error

Any input would help inmensly!
Best regards
 ­­
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The expression (1/((8 - 63^(1/2))^(1/2) + 1/((8 + 63^(1/2))^(1/2))^2 = 17 Feb 2024, 05:33
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ruis
Hi chetan2u, thank you for sharing your approach! I tried solving using identities as a first approach. Could you please help me understand what did I do wrong?­­
I then tried solving with your approach (multiplying by conjugate) and I am still getting it wrong. Can´t see where I commited the errors in both cases...

All input would help inmensly!
Best regards
 ­
­\((\frac{1}{\sqrt{8 - \sqrt{63}}} +\frac{1}{\sqrt{8 +\sqrt{63}}})^2 =\) has been taken as \((\frac{1}{\sqrt{8} - \sqrt{63}} +\frac{1}{\sqrt{8} +\sqrt{63}})^2 =\)

Entire denominator \(8-\sqrt{3}\) is under another square root.

Change and solve, and let me know if you don't get correct answer.
 
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The expression (1/((8 - 63^(1/2))^(1/2) + 1/((8 + 63^(1/2))^(1/2))^2 = 17 Feb 2024, 05:49
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chetan2u
ruis
Hi chetan2u, thank you for sharing your approach! I tried solving using identities as a first approach. Could you please help me understand what did I do wrong?­­
I then tried solving with your approach (multiplying by conjugate) and I am still getting it wrong. Can´t see where I commited the errors in both cases...

All input would help inmensly!
Best regards
 ­
­\((\frac{1}{\sqrt{8 - \sqrt{63}}} +\frac{1}{\sqrt{8 +\sqrt{63}}})^2 =\) has been taken as \((\frac{1}{\sqrt{8} - \sqrt{63}} +\frac{1}{\sqrt{8} +\sqrt{63}})^2 =\)

Entire denominator \(8-\sqrt{3}\) is under another square root.

Change and solve, and let me know if you don't get correct answer.

 
­Horrible careless error I commited there... Tried again but I am struggling with the last bit here 2 * a* b results in +2 and I do not see how.

Huge thanks for reponding me and helping me. Truly appreciate it
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Posts: 11,294
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 [1]
The expression (1/((8 - 63^(1/2))^(1/2) + 1/((8 + 63^(1/2))^(1/2))^2 = 17 Feb 2024, 05:57
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ruis
chetan2u
ruis
Hi chetan2u, thank you for sharing your approach! I tried solving using identities as a first approach. Could you please help me understand what did I do wrong?­­
I then tried solving with your approach (multiplying by conjugate) and I am still getting it wrong. Can´t see where I commited the errors in both cases...

All input would help inmensly!
Best regards
 ­
­\((\frac{1}{\sqrt{8 - \sqrt{63}}} +\frac{1}{\sqrt{8 +\sqrt{63}}})^2 =\) has been taken as \((\frac{1}{\sqrt{8} - \sqrt{63}} +\frac{1}{\sqrt{8} +\sqrt{63}})^2 =\)

Entire denominator \(8-\sqrt{3}\) is under another square root.

Change and solve, and let me know if you don't get correct answer.


 
­Horrible careless error I commited there... Tried again but I am struggling with the last bit here 2 * a* b results in +2 and I do not see how.

Huge thanks for reponding me and helping me. Truly appreciate it
­Another error is taking the sign as NEGATIVE while it is POSITIVE.
­\((\frac{1}{\sqrt{8 - \sqrt{63}}} +\frac{1}{\sqrt{8 +\sqrt{63}}})^2 \) has been taken as ­\((\frac{1}{\sqrt{8 - \sqrt{63}}} -\frac{1}{\sqrt{8 +\sqrt{63}}})^2 \)
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