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Question Number 82191 by jagoll last updated on 19/Feb/20
find the solution (√(x^2 −3x−4 )) > x−2
$${find}\:{the}\:{solution} \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:}\:\:>\:\:{x}−\mathrm{2}\: \\ $$
Commented by arkanmath7@gmail.com last updated on 19/Feb/20
x^2 −3x−4 > x^2 −4x+4 −3x−4 > −4x+4 x > 8 S.S. = {x:x>8} Or S.S.={(8,∞)} so easy
$${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:\:\:>\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4} \\ $$$$−\mathrm{3}{x}−\mathrm{4}\:\:\:>\:\:−\mathrm{4}{x}+\mathrm{4} \\ $$$${x}\:\:\:>\:\:\mathrm{8} \\ $$$${S}.{S}.\:=\:\left\{{x}:{x}>\mathrm{8}\right\}\: \\ $$$${Or}\:{S}.{S}.=\left\{\left(\mathrm{8},\infty\right)\right\} \\ $$$${so}\:{easy} \\ $$
Commented by jagoll last updated on 19/Feb/20
tes for x = −1 (√(1+3−4)) > −3 0 > −3 is correct sir
$${tes}\:{for}\:{x}\:=\:−\mathrm{1}\: \\ $$$$\sqrt{\mathrm{1}+\mathrm{3}−\mathrm{4}}\:>\:−\mathrm{3}\: \\ $$$$\mathrm{0}\:>\:−\mathrm{3}\:{is}\:{correct}\:{sir} \\ $$
Commented by jagoll last updated on 19/Feb/20
x = −1 ∉ (8,∞) but x=−1 is solution
$${x}\:=\:−\mathrm{1}\:\notin\:\left(\mathrm{8},\infty\right)\:{but}\:{x}=−\mathrm{1}\:{is}\:{solution} \\ $$
Commented by mr W last updated on 19/Feb/20
(√((x−4)(x+1)))>x−2 solution: x≤−1 ∨ x>8
$$\sqrt{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}>{x}−\mathrm{2} \\ $$$${solution}: \\ $$$${x}\leqslant−\mathrm{1}\:\vee\:{x}>\mathrm{8} \\ $$
Commented by arkanmath7@gmail.com last updated on 19/Feb/20
I think that′s true if you said (√(x^2 −3x−4)) ≥0 not when (√(x^2 −3x−4)) >x−2 put x=−1 in the 2nd line of the solution you will get a rong statement
$${I}\:{think}\:{that}'{s}\:{true}\:{if}\:{you}\:{said}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\geqslant\mathrm{0} \\ $$$${not}\:{when}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:>{x}−\mathrm{2} \\ $$$${put}\:{x}=−\mathrm{1}\:{in}\:{the}\:\mathrm{2}{nd}\:{line}\:{of}\:{the}\:{solution} \\ $$$${you}\:{will}\:{get}\:{a}\:{rong}\:{statement} \\ $$$$ \\ $$
Commented by Kunal12588 last updated on 19/Feb/20
x > 8 and (√((x+1)(x−4)))>(x−2) ⇒(((x+1)(x−4))/((x−2)^2 ))>0 ⇒x<−1
$${x}\:>\:\mathrm{8}\: \\ $$$${and}\:\sqrt{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{4}\right)}>\left({x}−\mathrm{2}\right) \\ $$$$\Rightarrow\frac{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{4}\right)}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$$$\Rightarrow{x}<−\mathrm{1} \\ $$
Commented by mr W last updated on 19/Feb/20
why use if (√(x^2 −3x−4))≥0 ? it is always true that (√(x^2 −3x−4))≥0.
$${why}\:{use}\:{if}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\geqslant\mathrm{0}\:?\:{it}\:{is}\:{always} \\ $$$${true}\:{that}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\geqslant\mathrm{0}. \\ $$
Commented by jagoll last updated on 19/Feb/20
so what is the answer conclusion? x≤−1 ∨x> 8 or x > 8?
$${so}\:{what}\:{is}\:{the}\:{answer}\:{conclusion}? \\ $$$${x}\leqslant−\mathrm{1}\:\vee{x}>\:\mathrm{8}\: \\ $$$${or}\:{x}\:>\:\mathrm{8}? \\ $$
Commented by mr W last updated on 19/Feb/20
when x≤−1, x−2≤−3, but (√(x^2 −3x−4))≥0, therefore (√(x^2 −3x−4))>x−2 is true. and x=−1 is also included.
$${when}\:{x}\leqslant−\mathrm{1},\:{x}−\mathrm{2}\leqslant−\mathrm{3},\:{but}\: \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\geqslant\mathrm{0},\:{therefore}\: \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}>{x}−\mathrm{2}\:{is}\:{true}.\:{and}\:{x}=−\mathrm{1} \\ $$$${is}\:{also}\:{included}. \\ $$
Commented by mr W last updated on 19/Feb/20
the result is x≤−1 ∨ x> 8 that means x∈ (−∞,−1] or x∈(8,+∞)
$${the}\:{result}\:{is} \\ $$$${x}\leqslant−\mathrm{1}\:\vee\:{x}>\:\mathrm{8}\: \\ $$$${that}\:{means}\:{x}\in\:\left(−\infty,−\mathrm{1}\right]\:{or}\:{x}\in\left(\mathrm{8},+\infty\right) \\ $$
Commented by jagoll last updated on 19/Feb/20
yes sir. i agree. but my teacher blamed my answer
$${yes}\:{sir}.\:{i}\:{agree}.\:{but}\:{my}\:{teacher}\: \\ $$$${blamed}\:{my}\:{answer} \\ $$
Commented by arkanmath7@gmail.com last updated on 19/Feb/20
confused quest i think you are the truest
$$ \\ $$$${confused}\:{quest} \\ $$$${i}\:{think}\:{you}\:{are}\:{the}\:{truest} \\ $$
Commented by jagoll last updated on 19/Feb/20
haha...if the most correct , surely God sir
$${haha}…{if}\:{the}\:{most}\:{correct}\:,\: \\ $$$${surely}\:{God}\:{sir} \\ $$
Commented by mr W last updated on 19/Feb/20
be very careful with squaring by inequlity! it may change the validity range of the original inequality!
$${be}\:{very}\:{careful}\:{with}\:{squaring}\:{by} \\ $$$${inequlity}!\:{it}\:{may}\:{change}\:{the}\:{validity} \\ $$$${range}\:{of}\:{the}\:{original}\:{inequality}! \\ $$
Commented by jagoll last updated on 19/Feb/20
please sir post your step solution
$${please}\:{sir}\:{post}\:{your}\:{step}\:{solution} \\ $$
Commented by jagoll last updated on 19/Feb/20
Commented by jagoll last updated on 19/Feb/20
it sir by graphic
$${it}\:{sir}\:{by}\:{graphic} \\ $$
Commented by mathmax 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,+∞[
$${the}\:{inequation}\:{is}\:{defined}\:\:{for}\:{x}\geqslant\mathrm{2}\:{and}\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\geqslant\mathrm{0} \\ $$$$\Delta=\left(−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{4}\right)=\mathrm{9}+\mathrm{16}=\mathrm{25}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{5}}{\mathrm{2}}=\mathrm{4}\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{5}}{\mathrm{2}}=−\mathrm{1} \\ $$$$\left.{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\geqslant\mathrm{0}\:\Rightarrow{x}\in\right]−\infty,−\mathrm{1}\left[\cup\right]\mathrm{4},+\infty\left[\:\Rightarrow\:{D}_{{in}} =\left[\mathrm{2},+\infty\left[\right.\right.\right. \\ $$$$\left({in}\right)\Rightarrow{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}>{x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{4}\:\Rightarrow{x}>\mathrm{8}\:\Rightarrow{S}\:=\left[\mathrm{8},+\infty\left[\right.\right. \\ $$
Commented by jagoll last updated on 19/Feb/20
solution x≤−1 ∨ x>8
$${solution}\:{x}\leqslant−\mathrm{1}\:\vee\:{x}>\mathrm{8} \\ $$
Answered by mr W last updated on 19/Feb/20
such that (√(x^2 −3x−4)) is defined, x^2 −3x−4=(x+1)(x−4)≥0 ⇒x≤−1 or x≥4 case 1: x≤−1 with x≤−1, we have x−2≤−3<0 but (√(x^2 −3x−4))≥0, so (√(x^2 −3x−4 )) > x−2 is true. ⇒x≤−1 is solution. case 2: x≥4 with x≥4, x−2≥6>0 x^2 −3x−4 >(x−2)^2 x^2 −3x−4 >x^2 −4x+4 x>8 ≥4 ⇒x>8 is solution. (we can use squaring here because we have ensured at first that x−2>0) summary of all solutions: x≤−1 x>8
$${such}\:{that}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:{is}\:{defined}, \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}=\left({x}+\mathrm{1}\right)\left({x}−\mathrm{4}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{x}\leqslant−\mathrm{1}\:{or}\:{x}\geqslant\mathrm{4} \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:{x}\leqslant−\mathrm{1} \\ $$$${with}\:{x}\leqslant−\mathrm{1},\:{we}\:{have}\:{x}−\mathrm{2}\leqslant−\mathrm{3}<\mathrm{0} \\ $$$${but}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\geqslant\mathrm{0},\:{so} \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:}\:>\:\:{x}−\mathrm{2}\:\:{is}\:{true}. \\ $$$$\Rightarrow{x}\leqslant−\mathrm{1}\:{is}\:{solution}. \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{x}\geqslant\mathrm{4} \\ $$$${with}\:{x}\geqslant\mathrm{4},\:{x}−\mathrm{2}\geqslant\mathrm{6}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:>\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:>{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4} \\ $$$${x}>\mathrm{8}\:\geqslant\mathrm{4} \\ $$$$\Rightarrow{x}>\mathrm{8}\:{is}\:{solution}. \\ $$$$\left({we}\:{can}\:{use}\:{squaring}\:{here}\:{because}\:{we}\right. \\ $$$$\left.{have}\:{ensured}\:{at}\:{first}\:{that}\:{x}−\mathrm{2}>\mathrm{0}\right) \\ $$$$ \\ $$$${summary}\:{of}\:{all}\:{solutions}: \\ $$$${x}\leqslant−\mathrm{1} \\ $$$${x}>\mathrm{8} \\ $$
Commented by jagoll last updated on 19/Feb/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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