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Question Number 79279 by jagoll last updated on 24/Jan/20

solve ∣x∣^3 −7x^2 +7∣x∣+15<0

$$\mathrm{solve}\: \\ $$ $$\mid\mathrm{x}\mid^{\mathrm{3}} −\mathrm{7x}^{\mathrm{2}} +\mathrm{7}\mid\mathrm{x}\mid+\mathrm{15}<\mathrm{0} \\ $$

Commented byjagoll last updated on 24/Jan/20

thanks

$$\mathrm{thanks}\: \\ $$

Commented byMJS last updated on 24/Jan/20

sorry but it′s wrong

$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$

Commented byjohn santu last updated on 24/Jan/20

where is wrong?

$${where}\:{is}\:{wrong}? \\ $$

Commented byMJS last updated on 24/Jan/20

you should plot the function or scetch it

$$\mathrm{you}\:\mathrm{should}\:\mathrm{plot}\:\mathrm{the}\:\mathrm{function}\:\mathrm{or}\:\mathrm{scetch}\:\mathrm{it} \\ $$

Answered by MJS last updated on 24/Jan/20

case 1 x≥0 x^3 −7x^2 +7x+15=0 ⇒ x=−1∨x=3∨x=5 testing x=4: 4^3 −7×4^2 +7×4+15=−5 case 2 x<0 −x^3 −7x^2 −7x+15=0 ⇒ x=−5∨x=−3∨x=1 testing x=−4: −(−4)^3 −7×(−4)^2 −7×4+15=−5 ⇒ −5<x<−3∨3<x<5 or 3<∣x∣<5

$$\mathrm{case}\:\mathrm{1}\:{x}\geqslant\mathrm{0} \\ $$ $${x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{15}=\mathrm{0} \\ $$ $$\Rightarrow\:{x}=−\mathrm{1}\vee{x}=\mathrm{3}\vee{x}=\mathrm{5} \\ $$ $$\mathrm{testing}\:{x}=\mathrm{4}:\:\mathrm{4}^{\mathrm{3}} −\mathrm{7}×\mathrm{4}^{\mathrm{2}} +\mathrm{7}×\mathrm{4}+\mathrm{15}=−\mathrm{5} \\ $$ $$\mathrm{case}\:\mathrm{2}\:{x}<\mathrm{0} \\ $$ $$−{x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{15}=\mathrm{0} \\ $$ $$\Rightarrow\:{x}=−\mathrm{5}\vee{x}=−\mathrm{3}\vee{x}=\mathrm{1} \\ $$ $$\mathrm{testing}\:{x}=−\mathrm{4}:\:−\left(−\mathrm{4}\right)^{\mathrm{3}} −\mathrm{7}×\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{7}×\mathrm{4}+\mathrm{15}=−\mathrm{5} \\ $$ $$ \\ $$ $$\Rightarrow \\ $$ $$−\mathrm{5}<{x}<−\mathrm{3}\vee\mathrm{3}<{x}<\mathrm{5} \\ $$ $$\mathrm{or} \\ $$ $$\mathrm{3}<\mid{x}\mid<\mathrm{5} \\ $$

Commented byjohn santu last updated on 24/Jan/20

your are right. i′m forgot the definition sign(x)

$${your}\:{are}\:{right}.\:{i}'{m}\:{forgot}\: \\ $$ $${the}\:{definition}\:{sign}\left({x}\right) \\ $$

Commented byjagoll last updated on 24/Jan/20

thanks you sir

$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

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