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Question Number 96155 by pticantor last updated on 30/May/20

Commented by prakash jain last updated on 30/May/20

$$c=\sqrt{4-\sqrt{5+c}}$$$$c^2=4-\sqrt{5+c}$$$$c^2-4=-\sqrt{5+c}$$$$c^4-8c^2+16=5+c$$$$c^4-8c^2-c+11=0$$$$\mathrm{Hence}\mathrm{c}\mathrm{is}\mathrm{a}\mathrm{solution}\mathrm{to}$$$$y^4-8y^2-y+11=0$$

Commented by prakash jain last updated on 30/May/20

$$eqn1:$$$$x^4-8x^2+x+11=0\left(\mathrm{i}\right)$$$$y^4-8y^2-y+11=0\left(\mathrm{ii}\right)$$$$\mathrm{if}\alpha \mathrm{is}\mathrm{a}\mathrm{root}\mathrm{of}\mathrm{equation}\left(\mathrm{ii}\right)$$$$\mathrm{then}\mathrm{we}\mathrm{can}\mathrm{see}(-\alpha )\mathrm{will}\mathrm{be}$$$$\mathrm{root}\mathrm{of}\mathrm{equation}\left(\mathrm{i}\right).$$$$\mathrm{so}\mathrm{eqn}\left(\mathrm{i}\right)\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}$$$$(x-a)(x-b)(x+c)(x+d)=0$$$$\mathrm{comparing}\mathrm{coeffcient}\mathrm{of}\mathrm{constant}$$$$\mathrm{term}$$$$abcd=11$$

Commented by mr W last updated on 30/May/20

$$nicesolution!$$

Commented by JDamian last updated on 30/May/20

Why do you repeat the same questions? this one is the same Q95967

Answered by pticantor last updated on 30/May/20

$$pleaseineedhelponquestion3>>$$

Commented by prakash jain last updated on 30/May/20

$$\mathrm{Please}\mathrm{see}\mathrm{comments}.$$$$\mathrm{I}\mathrm{think}abcd=11$$

Commented by pticantor last updated on 30/May/20

$${it}^{\prime }scorrect\mathit{s}\mathit{i}\mathit{r}\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{t}\mathit{o}\mathit{o}\mathit{m}\mathit{u}\mathit{c}\mathit{h}$$

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