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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
Howcansuchquestionsbesolved.?x2−∣7∣+10=0x2−∣x∣−6>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
x2−∣x∣−6=0case1:x<0⇒∣x∣=−xx2+x−6=0⇒x=−3∨x=2butx<0⇒x=−3case2:x⩾0⇒∣x∣=xx2−x−6=0⇒x=−2∨x=3butx⩾0⇒x=3x2−∣x∣−6=0⇒x=±3x2−∣x∣−6>0⇒x<−3∨x>3
Commented by Rio Michael last updated on 12/Jul/19
does it also apply to x^2 +∣7x∣+10=0?
doesitalsoapplytox2+∣7x∣+10=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
x2+∣7x∣+10=0x2+7∣x∣+10=0case1x<0⇒∣x∣=−xx2−7x+10=0⇒x=2∨x=5butx<0⇒nosolutioncase2x⩾0⇒∣x∣=xx2+7x+10=0⇒x=−2∨x=−5butx⩾0⇒nosolutiontheabsoluteminimumofx2+7∣x∣+10is10
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
wemightgetcomplexsolutionsx=a+bileadsto(a+bi)2+7∣a+bi∣+10=0a2−b2+7a2+b2+10+2abi=0⇒a2−b2+7a2+b2+10=0∧2abi=02abi=0⇒a=0∨b=0a=0−b2+7∣b∣+10=0⇒again2cases⇒⇒b=−72−892∨b=72+892⇒x=±(72+892)ib=0leadstotheaboveequationonlywithainsteadofx⇒norealsolutions
Commented by Rio Michael last updated on 12/Jul/19
thank you
thankyou

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