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Author: Tinku Tara

suppose-that-E-9-4-k-2-n-a-k-2-is-given-let-a-k-a-k-1-a-k-1-1-and-a-1-1-a-2-2-a-3-6-

Question Number 198813 by mnjuly1970 last updated on 24/Oct/23 $$ \\ $$$$\:\:\:\:\:{suppose}\:\:{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:{E}=\:\sqrt{\:\mathrm{9}\:+\:\mathrm{4}\:\left(\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\:\:{a}_{\:{k}} ^{\:\mathrm{2}} \:\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{is}\:\:{given} \\ $$$$\:\:\:\:\:{let}\:\:\:\:{a}_{\:{k}} \:=\:{a}_{{k}−\mathrm{1}}…

Simplify-x-4-x-4-x-4-x-4-x-8-4-x-2-12x-32-

Question Number 198814 by hardmath last updated on 24/Oct/23 $$\mathrm{Simplify}: \\ $$$$\frac{\left(\sqrt{\mathrm{x}\:+\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\:+\:\sqrt{\mathrm{x}\:-\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\right)\:\sqrt{\mathrm{x}\:-\:\mathrm{8}}}{\mathrm{4}\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{12x}\:+\:\mathrm{32}}} \\ $$ Answered by a.lgnaoui last updated on 24/Oct/23 $$=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\mathrm{x}+\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}{\boldsymbol{\mathrm{x}}−\mathrm{4}}}\:+\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\boldsymbol{\mathrm{x}}−\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}{\boldsymbol{\mathrm{x}}−\mathrm{4}}} \\ $$$$=\frac{\sqrt{\boldsymbol{\mathrm{x}}+\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}\:}\:+\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}}{\:\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}…

Prove-the-following-is-a-tautology-p-q-p-r-q-r-

Question Number 198750 by depressiveshrek last updated on 24/Oct/23 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{a}\:\mathrm{tautology}: \\ $$$$\left[\left({p}\veebar{q}\right)\wedge\left({p}\Rightarrow{r}\right)\right]\Rightarrow\left({q}\veebar{r}\right) \\ $$ Answered by MathematicalUser2357 last updated on 29/Dec/23 $$\mathrm{Only}\:\mathrm{know}\:\mathrm{until} \\ $$$$=\sim\left[\left({p}\veebar{q}\right)\wedge\left(\sim{p}\vee{r}\right)\right]\vee\left({q}\veebar{r}\right) \\…

x-4-ax-3-bx-2-cx-d-0-

Question Number 198772 by ajfour last updated on 24/Oct/23 $${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$ Commented by Frix last updated on 24/Oct/23 $$\mathrm{Usually}\:\mathrm{we}\:\mathrm{first}\:\mathrm{try}\:\mathrm{factors}\:\mathrm{of}\:{d}.\:\mathrm{The}\:\mathrm{next} \\ $$$$\mathrm{step}\:\mathrm{would}\:\mathrm{be}\:\mathrm{substituting}\:{x}={t}−\frac{{a}}{\mathrm{4}}\:\mathrm{to}…