Question Number 82191 by jagoll last updated on 19/Feb/20
$${find}\:{the}\:{solution} \\ $$ $$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:}\:\:>\:\:{x}−\mathrm{2}\: \\ $$
Commented byarkanmath7@gmail.com last updated on 19/Feb/20
$${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:\:\:>\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4} \\ $$ $$−\mathrm{3}{x}−\mathrm{4}\:\:\:>\:\:−\mathrm{4}{x}+\mathrm{4} \\ $$ $${x}\:\:\:>\:\:\mathrm{8} \\ $$ $${S}.{S}.\:=\:\left\{{x}:{x}>\mathrm{8}\right\}\: \\ $$ $${Or}\:{S}.{S}.=\left\{\left(\mathrm{8},\infty\right)\right\} \\ $$ $${so}\:{easy} \\ $$
Commented byjagoll last updated on 19/Feb/20
$${tes}\:{for}\:{x}\:=\:−\mathrm{1}\: \\ $$ $$\sqrt{\mathrm{1}+\mathrm{3}−\mathrm{4}}\:>\:−\mathrm{3}\: \\ $$ $$\mathrm{0}\:>\:−\mathrm{3}\:{is}\:{correct}\:{sir} \\ $$
Commented byjagoll last updated on 19/Feb/20
$${x}\:=\:−\mathrm{1}\:\notin\:\left(\mathrm{8},\infty\right)\:{but}\:{x}=−\mathrm{1}\:{is}\:{solution} \\ $$
Commented bymr W last updated on 19/Feb/20
$$\sqrt{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}>{x}−\mathrm{2} \\ $$ $${solution}: \\ $$ $${x}\leqslant−\mathrm{1}\:\vee\:{x}>\mathrm{8} \\ $$
Commented byarkanmath7@gmail.com last updated on 19/Feb/20
$${I}\:{think}\:{that}'{s}\:{true}\:{if}\:{you}\:{said}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\geqslant\mathrm{0} \\ $$ $${not}\:{when}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:>{x}−\mathrm{2} \\ $$ $${put}\:{x}=−\mathrm{1}\:{in}\:{the}\:\mathrm{2}{nd}\:{line}\:{of}\:{the}\:{solution} \\ $$ $${you}\:{will}\:{get}\:{a}\:{rong}\:{statement} \\ $$ $$ \\ $$
Commented byKunal12588 last updated on 19/Feb/20
$${x}\:>\:\mathrm{8}\: \\ $$ $${and}\:\sqrt{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{4}\right)}>\left({x}−\mathrm{2}\right) \\ $$ $$\Rightarrow\frac{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{4}\right)}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$ $$\Rightarrow{x}<−\mathrm{1} \\ $$
Commented bymr W last updated on 19/Feb/20
$${why}\:{use}\:{if}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\geqslant\mathrm{0}\:?\:{it}\:{is}\:{always} \\ $$ $${true}\:{that}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\geqslant\mathrm{0}. \\ $$
Commented byjagoll last updated on 19/Feb/20
$${so}\:{what}\:{is}\:{the}\:{answer}\:{conclusion}? \\ $$ $${x}\leqslant−\mathrm{1}\:\vee{x}>\:\mathrm{8}\: \\ $$ $${or}\:{x}\:>\:\mathrm{8}? \\ $$
Commented bymr W last updated on 19/Feb/20
$${when}\:{x}\leqslant−\mathrm{1},\:{x}−\mathrm{2}\leqslant−\mathrm{3},\:{but}\: \\ $$ $$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\geqslant\mathrm{0},\:{therefore}\: \\ $$ $$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}>{x}−\mathrm{2}\:{is}\:{true}.\:{and}\:{x}=−\mathrm{1} \\ $$ $${is}\:{also}\:{included}. \\ $$
Commented bymr W last updated on 19/Feb/20
![the result is x≤−1 ∨ x> 8 that means x∈ (−∞,−1] or x∈(8,+∞)](Q82204.png)
$${the}\:{result}\:{is} \\ $$ $${x}\leqslant−\mathrm{1}\:\vee\:{x}>\:\mathrm{8}\: \\ $$ $${that}\:{means}\:{x}\in\:\left(−\infty,−\mathrm{1}\right]\:{or}\:{x}\in\left(\mathrm{8},+\infty\right) \\ $$
Commented byjagoll last updated on 19/Feb/20
$${yes}\:{sir}.\:{i}\:{agree}.\:{but}\:{my}\:{teacher}\: \\ $$ $${blamed}\:{my}\:{answer} \\ $$
Commented byarkanmath7@gmail.com last updated on 19/Feb/20
$$ \\ $$ $${confused}\:{quest} \\ $$ $${i}\:{think}\:{you}\:{are}\:{the}\:{truest} \\ $$
Commented byjagoll last updated on 19/Feb/20
$${haha}...{if}\:{the}\:{most}\:{correct}\:,\: \\ $$ $${surely}\:{God}\:{sir} \\ $$
Commented bymr W last updated on 19/Feb/20
$${be}\:{very}\:{careful}\:{with}\:{squaring}\:{by} \\ $$ $${inequlity}!\:{it}\:{may}\:{change}\:{the}\:{validity} \\ $$ $${range}\:{of}\:{the}\:{original}\:{inequality}! \\ $$
Commented byjagoll last updated on 19/Feb/20
$${please}\:{sir}\:{post}\:{your}\:{step}\:{solution} \\ $$
Commented byjagoll last updated on 19/Feb/20
Commented byjagoll last updated on 19/Feb/20
$${it}\:{sir}\:{by}\:{graphic} \\ $$
Commented bymathmax by abdo last updated on 19/Feb/20
![the inequation is defined for x≥2 and x^2 −3x−4≥0 Δ=(−3)^2 −4(−4)=9+16=25 ⇒x_1 =((3+5)/2)=4 and x_2 =((3−5)/2)=−1 x^2 −3x−4≥0 ⇒x∈]−∞,−1[∪]4,+∞[ ⇒ D_(in) =[2,+∞[ (in)⇒x^2 −3x−4>x^2 −4x +4 ⇒x>8 ⇒S =[8,+∞[](Q82227.png)
$${the}\:{inequation}\:{is}\:{defined}\:\:{for}\:{x}\geqslant\mathrm{2}\:{and}\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\geqslant\mathrm{0} \\ $$ $$\Delta=\left(−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{4}\right)=\mathrm{9}+\mathrm{16}=\mathrm{25}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{5}}{\mathrm{2}}=\mathrm{4}\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{5}}{\mathrm{2}}=−\mathrm{1} \\ $$ $$\left.{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\geqslant\mathrm{0}\:\Rightarrow{x}\in\right]−\infty,−\mathrm{1}\left[\cup\right]\mathrm{4},+\infty\left[\:\Rightarrow\:{D}_{{in}} =\left[\mathrm{2},+\infty\left[\right.\right.\right. \\ $$ $$\left({in}\right)\Rightarrow{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}>{x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{4}\:\Rightarrow{x}>\mathrm{8}\:\Rightarrow{S}\:=\left[\mathrm{8},+\infty\left[\right.\right. \\ $$
Commented byjagoll last updated on 19/Feb/20
$${solution}\:{x}\leqslant−\mathrm{1}\:\vee\:{x}>\mathrm{8} \\ $$
Answered by mr W last updated on 19/Feb/20
$${such}\:{that}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:{is}\:{defined}, \\ $$ $${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}=\left({x}+\mathrm{1}\right)\left({x}−\mathrm{4}\right)\geqslant\mathrm{0} \\ $$ $$\Rightarrow{x}\leqslant−\mathrm{1}\:{or}\:{x}\geqslant\mathrm{4} \\ $$ $$ \\ $$ $${case}\:\mathrm{1}:\:{x}\leqslant−\mathrm{1} \\ $$ $${with}\:{x}\leqslant−\mathrm{1},\:{we}\:{have}\:{x}−\mathrm{2}\leqslant−\mathrm{3}<\mathrm{0} \\ $$ $${but}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\geqslant\mathrm{0},\:{so} \\ $$ $$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:}\:>\:\:{x}−\mathrm{2}\:\:{is}\:{true}. \\ $$ $$\Rightarrow{x}\leqslant−\mathrm{1}\:{is}\:{solution}. \\ $$ $$ \\ $$ $${case}\:\mathrm{2}:\:{x}\geqslant\mathrm{4} \\ $$ $${with}\:{x}\geqslant\mathrm{4},\:{x}−\mathrm{2}\geqslant\mathrm{6}>\mathrm{0} \\ $$ $${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:>\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$ $${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:>{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4} \\ $$ $${x}>\mathrm{8}\:\geqslant\mathrm{4} \\ $$ $$\Rightarrow{x}>\mathrm{8}\:{is}\:{solution}. \\ $$ $$\left({we}\:{can}\:{use}\:{squaring}\:{here}\:{because}\:{we}\right. \\ $$ $$\left.{have}\:{ensured}\:{at}\:{first}\:{that}\:{x}−\mathrm{2}>\mathrm{0}\right) \\ $$ $$ \\ $$ $${summary}\:{of}\:{all}\:{solutions}: \\ $$ $${x}\leqslant−\mathrm{1} \\ $$ $${x}>\mathrm{8} \\ $$
Commented byjagoll last updated on 19/Feb/20
$${thank}\:{you}\:{sir} \\ $$