Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 733 by 123456 last updated on 05/Mar/15

∣∣∣∣∣∣∣x^2 −x−1∣−3∣−5∣−7∣−9∣−11∣−13∣ =x^2 −2x−48 find all x real that is solution of above equation

$$\mid\mid\mid\mid\mid\mid\mid{x}^{\mathrm{2}} −{x}−\mathrm{1}\mid−\mathrm{3}\mid−\mathrm{5}\mid−\mathrm{7}\mid−\mathrm{9}\mid−\mathrm{11}\mid−\mathrm{13}\mid \\ $$$$={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{48} \\ $$$${find}\:{all}\:{x}\:{real}\:{that}\:{is}\:{solution}\:{of}\:{above} \\ $$$${equation} \\ $$

Answered by prakash jain last updated on 05/Mar/15

x^2 −2x−48=(x−8)(x+6) (x−8)(x+6)≥0 for x≥8 or x≤−6 Case I: x≥8 x^2 −x−1>0 ... x^2 −x−49>0 x^2 −x−49=x^2 −2x−48 x=1 ⇒ No solution for x≥8 Case II: x≤−6 x^2 −x−1>0 ... x^2 −x−36>0 x^2 −x−49=(x−((1+(√(197)))/2))(x−((1−(√(197)))/2)) x^2 −x−49≥0 gives only solution x=1 −6 x^2 −x−49<0 −x^2 +x+49=x^2 −2x−48 2x^2 −3x−97=0 Root for x≤−6 x=((3−(√(9+8×97)))/4)=((3−(√(9+776)))/4)=((3−(√(785)))/4) ((3−(√(785)))/4)−((1+(√(197)))/2)=((3−(√(785))−2−(√(788)))/4)<0 ((3−(√(785)))/4)−((1−(√(197)))/2)=((3−(√(785))−2+(√(788)))/4)>0 x^2 −x−49<0 for x=((3−(√(785)))/4) ((3−(√(785)))/4)+6=((27−(√(785)))/4)=(((√(729))−(√(785)))/4)<0 ((3−(√(785)))/4)<−6 and x^2 −x−49<0 for x=((3−(√(785)))/4) so it a valid solution. Solution for real x x=((3−(√(785)))/4)

$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{48}=\left({x}−\mathrm{8}\right)\left({x}+\mathrm{6}\right) \\ $$$$\left({x}−\mathrm{8}\right)\left({x}+\mathrm{6}\right)\geqslant\mathrm{0}\:\mathrm{for}\:{x}\geqslant\mathrm{8}\:\mathrm{or}\:{x}\leqslant−\mathrm{6} \\ $$$$\mathrm{Case}\:\mathrm{I}:\:{x}\geqslant\mathrm{8} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}>\mathrm{0} \\ $$$$... \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{49}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{49}={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{48} \\ $$$${x}=\mathrm{1}\:\Rightarrow\:\mathrm{No}\:\mathrm{solution}\:\mathrm{for}\:{x}\geqslant\mathrm{8} \\ $$$$\mathrm{Case}\:\mathrm{II}:\:{x}\leqslant−\mathrm{6} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}>\mathrm{0} \\ $$$$... \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{36}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{49}=\left({x}−\frac{\mathrm{1}+\sqrt{\mathrm{197}}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}−\sqrt{\mathrm{197}}}{\mathrm{2}}\right) \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{49}\geqslant\mathrm{0}\:\mathrm{gives}\:\mathrm{only}\:\mathrm{solution}\:{x}=\mathrm{1} −\mathrm{6} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{49}<\mathrm{0}\: \\ $$$$−{x}^{\mathrm{2}} +{x}+\mathrm{49}={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{48} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{97}=\mathrm{0} \\ $$$$\mathrm{Root}\:\mathrm{for}\:{x}\leqslant−\mathrm{6} \\ $$$${x}=\frac{\mathrm{3}−\sqrt{\mathrm{9}+\mathrm{8}×\mathrm{97}}}{\mathrm{4}}=\frac{\mathrm{3}−\sqrt{\mathrm{9}+\mathrm{776}}}{\mathrm{4}}=\frac{\mathrm{3}−\sqrt{\mathrm{785}}}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}−\sqrt{\mathrm{785}}}{\mathrm{4}}−\frac{\mathrm{1}+\sqrt{\mathrm{197}}}{\mathrm{2}}=\frac{\mathrm{3}−\sqrt{\mathrm{785}}−\mathrm{2}−\sqrt{\mathrm{788}}}{\mathrm{4}}<\mathrm{0} \\ $$$$\frac{\mathrm{3}−\sqrt{\mathrm{785}}}{\mathrm{4}}−\frac{\mathrm{1}−\sqrt{\mathrm{197}}}{\mathrm{2}}=\frac{\mathrm{3}−\sqrt{\mathrm{785}}−\mathrm{2}+\sqrt{\mathrm{788}}}{\mathrm{4}}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{49}<\mathrm{0}\:\mathrm{for}\:{x}=\frac{\mathrm{3}−\sqrt{\mathrm{785}}}{\mathrm{4}}\: \\ $$$$\frac{\mathrm{3}−\sqrt{\mathrm{785}}}{\mathrm{4}}+\mathrm{6}=\frac{\mathrm{27}−\sqrt{\mathrm{785}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{729}}−\sqrt{\mathrm{785}}}{\mathrm{4}}<\mathrm{0} \\ $$$$\frac{\mathrm{3}−\sqrt{\mathrm{785}}}{\mathrm{4}}<−\mathrm{6}\:\mathrm{and}\:{x}^{\mathrm{2}} −{x}−\mathrm{49}<\mathrm{0}\:\mathrm{for}\:{x}=\frac{\mathrm{3}−\sqrt{\mathrm{785}}}{\mathrm{4}} \\ $$$$\mathrm{so}\:\mathrm{it}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{solution}. \\ $$$$\mathrm{Solution}\:\mathrm{for}\:\mathrm{real}\:{x} \\ $$$${x}=\frac{\mathrm{3}−\sqrt{\mathrm{785}}}{\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com