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Question Number 64015 by Rio Michael last updated on 12/Jul/19
How can such questions be solved.? x^2 −∣7∣ +10=0 x^2 −∣x∣−6>0
$$\:{How}\:{can}\:{such}\:{questions}\:{be}\:{solved}.? \\ $$$$\:\:{x}^{\mathrm{2}} −\mid\mathrm{7}\mid\:+\mathrm{10}=\mathrm{0} \\ $$$$\:\:{x}^{\mathrm{2}} −\mid{x}\mid−\mathrm{6}>\mathrm{0} \\ $$$$\:\: \\ $$
Answered by MJS last updated on 12/Jul/19
x^2 −∣x∣−6=0 case 1: x<0 ⇒ ∣x∣=−x x^2 +x−6=0 ⇒ x=−3∨x=2 but x<0 ⇒ x=−3 case 2: x≥0 ⇒ ∣x∣=x x^2 −x−6=0 ⇒ x=−2∨x=3 but x≥0 ⇒ x=3 x^2 −∣x∣−6=0 ⇒ x=±3 x^2 −∣x∣−6>0 ⇒ x<−3∨x>3
$${x}^{\mathrm{2}} −\mid{x}\mid−\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{case}\:\mathrm{1}:\:{x}<\mathrm{0}\:\Rightarrow\:\mid{x}\mid=−{x} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{6}=\mathrm{0}\:\Rightarrow\:{x}=−\mathrm{3}\vee{x}=\mathrm{2}\:\mathrm{but}\:{x}<\mathrm{0}\:\Rightarrow\:{x}=−\mathrm{3} \\ $$$$\mathrm{case}\:\mathrm{2}:\:{x}\geqslant\mathrm{0}\:\Rightarrow\:\mid{x}\mid={x} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}=\mathrm{0}\:\Rightarrow\:{x}=−\mathrm{2}\vee{x}=\mathrm{3}\:\mathrm{but}\:{x}\geqslant\mathrm{0}\:\Rightarrow\:{x}=\mathrm{3} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mid{x}\mid−\mathrm{6}=\mathrm{0}\:\Rightarrow\:{x}=\pm\mathrm{3} \\ $$$${x}^{\mathrm{2}} −\mid{x}\mid−\mathrm{6}>\mathrm{0}\:\Rightarrow\:{x}<−\mathrm{3}\vee{x}>\mathrm{3} \\ $$
Commented by Rio Michael last updated on 12/Jul/19
does it also apply to x^2 +∣7x∣+10=0?
$${does}\:{it}\:{also}\:{apply}\:{to}\:{x}^{\mathrm{2}} +\mid\mathrm{7}{x}\mid+\mathrm{10}=\mathrm{0}? \\ $$
Commented by MJS last updated on 12/Jul/19
x^2 +∣7x∣+10=0 x^2 +7∣x∣+10=0 case 1 x<0 ⇒ ∣x∣=−x x^2 −7x+10=0 ⇒ x=2∨x=5 but x<0 ⇒ no solution case 2 x≥0 ⇒ ∣x∣=x x^2 +7x+10=0 ⇒ x=−2∨x=−5 but x≥0 ⇒ no solution the absolute minimum of x^2 +7∣x∣+10 is 10
$${x}^{\mathrm{2}} +\mid\mathrm{7}{x}\mid+\mathrm{10}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{7}\mid{x}\mid+\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{case}\:\mathrm{1}\:{x}<\mathrm{0}\:\Rightarrow\:\mid{x}\mid=−{x} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{10}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{2}\vee{x}=\mathrm{5}\:\mathrm{but}\:{x}<\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{case}\:\mathrm{2}\:{x}\geqslant\mathrm{0}\:\Rightarrow\:\mid{x}\mid={x} \\ $$$${x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{10}=\mathrm{0}\:\Rightarrow\:{x}=−\mathrm{2}\vee{x}=−\mathrm{5}\:\mathrm{but}\:{x}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{the}\:\mathrm{absolute}\:\mathrm{minimum}\:\mathrm{of}\:{x}^{\mathrm{2}} +\mathrm{7}\mid{x}\mid+\mathrm{10}\:\mathrm{is}\:\mathrm{10} \\ $$
Commented by MJS last updated on 12/Jul/19
we might get complex solutions x=a+bi leads to (a+bi)^2 +7∣a+bi∣+10=0 a^2 −b^2 +7(√(a^2 +b^2 ))+10+2abi=0 ⇒ a^2 −b^2 +7(√(a^2 +b^2 ))+10=0∧2abi=0 2abi=0 ⇒ a=0∨b=0 a=0 −b^2 +7∣b∣+10=0 ⇒ again 2 cases ⇒ ⇒ b=−(7/2)−((√(89))/2)∨b=(7/2)+((√(89))/2) ⇒ x=±((7/2)+((√(89))/2))i b=0 leads to the above equation only with a instead of x ⇒ no real solutions
$$\mathrm{we}\:\mathrm{might}\:\mathrm{get}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$${x}={a}+{b}\mathrm{i} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$$\left({a}+{b}\mathrm{i}\right)^{\mathrm{2}} +\mathrm{7}\mid{a}+{b}\mathrm{i}\mid+\mathrm{10}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{7}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\mathrm{10}+\mathrm{2}{ab}\mathrm{i}=\mathrm{0} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{7}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\mathrm{10}=\mathrm{0}\wedge\mathrm{2}{ab}\mathrm{i}=\mathrm{0} \\ $$$$\mathrm{2}{ab}\mathrm{i}=\mathrm{0}\:\Rightarrow\:{a}=\mathrm{0}\vee{b}=\mathrm{0} \\ $$$${a}=\mathrm{0} \\ $$$$−\mathrm{b}^{\mathrm{2}} +\mathrm{7}\mid{b}\mid+\mathrm{10}=\mathrm{0}\:\Rightarrow\:\mathrm{again}\:\mathrm{2}\:\mathrm{cases}\:\Rightarrow \\ $$$$\Rightarrow\:{b}=−\frac{\mathrm{7}}{\mathrm{2}}−\frac{\sqrt{\mathrm{89}}}{\mathrm{2}}\vee{b}=\frac{\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{\mathrm{89}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\pm\left(\frac{\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{\mathrm{89}}}{\mathrm{2}}\right)\mathrm{i} \\ $$$$ \\ $$$${b}=\mathrm{0}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{above}\:\mathrm{equation}\:\mathrm{only}\:\mathrm{with} \\ $$$${a}\:\mathrm{instead}\:\mathrm{of}\:{x}\:\Rightarrow\:\mathrm{no}\:\mathrm{real}\:\mathrm{solutions} \\ $$
Commented by Rio Michael last updated on 12/Jul/19
thank you
$${thank}\:{you} \\ $$

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