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Question Number 110399 by Aina Samuel Temidayo last updated on 29/Aug/20

The identity 2[16a^4 +81b^4 +c^4 ]=[4a^2 +9b^2 +c^2 ]^2 cannot result from which of the following equations? A. 6b=4a+2c B. 6a=9b+3c C. 6b=−4a+2c D.c= −2a−3b E. 6c=2b+3a

$$\mathrm{The}\:\mathrm{identity} \\ $$$$\mathrm{2}\left[\mathrm{16a}^{\mathrm{4}} +\mathrm{81b}^{\mathrm{4}} +\mathrm{c}^{\mathrm{4}} \right]=\left[\mathrm{4a}^{\mathrm{2}} +\mathrm{9b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right]^{\mathrm{2}} \\ $$$$\mathrm{cannot}\:\mathrm{result}\:\mathrm{from}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{equations}?\: \\ $$$$\mathrm{A}.\:\mathrm{6b}=\mathrm{4a}+\mathrm{2c}\:\mathrm{B}.\:\mathrm{6a}=\mathrm{9b}+\mathrm{3c}\:\mathrm{C}. \\ $$$$\mathrm{6b}=−\mathrm{4a}+\mathrm{2c}\:\mathrm{D}.\mathrm{c}=\:−\mathrm{2a}−\mathrm{3b}\:\mathrm{E}. \\ $$$$\mathrm{6c}=\mathrm{2b}+\mathrm{3a} \\ $$

Commented by Rasheed.Sindhi last updated on 28/Aug/20

D. −2a−3b

$$\mathrm{D}.\:−\mathrm{2a}−\mathrm{3b} \\ $$

Commented by Aina Samuel Temidayo last updated on 28/Aug/20

Your solution please? I think it′s E.

$$\mathrm{Your}\:\mathrm{solution}\:\mathrm{please}?\:\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{E}. \\ $$

Commented by 1549442205PVT last updated on 29/Aug/20

Your question wrote correctly D.−2a−3b isn′t an equation.If this is c=−2a−3b then answer will be E.

$$\mathrm{Your}\:\mathrm{question}\:\mathrm{wrote}\:\mathrm{correctly} \\ $$$$\boldsymbol{\mathrm{D}}.−\mathrm{2}\boldsymbol{\mathrm{a}}−\mathrm{3}\boldsymbol{\mathrm{b}}\:\boldsymbol{\mathrm{isn}}'\boldsymbol{\mathrm{t}}\:\boldsymbol{\mathrm{a}}\mathrm{n}\:\boldsymbol{\mathrm{equation}}.\mathrm{If}\:\mathrm{this}\: \\ $$$$\mathrm{is}\:\boldsymbol{\mathrm{c}}=−\mathrm{2}\boldsymbol{\mathrm{a}}−\mathrm{3}\boldsymbol{\mathrm{b}}\:\mathrm{then}\:\mathrm{answer}\:\mathrm{will}\:\mathrm{be} \\ $$$$\mathrm{E}. \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Oh. Sorry, that was a mistake from me. It has been corrected. So, how did you get the answer to be E?

$$\mathrm{Oh}.\:\mathrm{Sorry},\:\mathrm{that}\:\mathrm{was}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{from} \\ $$$$\mathrm{me}.\:\mathrm{It}\:\mathrm{has}\:\mathrm{been}\:\mathrm{corrected}.\:\mathrm{So},\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{to} \\ $$$$\mathrm{be}\:\mathrm{E}? \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

The question has been corrected. You can check again. Thanks.

$$\mathrm{The}\:\mathrm{question}\:\mathrm{has}\:\mathrm{been}\:\mathrm{corrected}.\:\mathrm{You} \\ $$$$\mathrm{can}\:\mathrm{check}\:\mathrm{again}.\:\mathrm{Thanks}. \\ $$

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