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The-real-roots-of-the-equation-x-2-6x-c-0-differ-by-2n-where-n-is-a-real-non-zero-Show-that-n-2-9-c-Given-that-the-roots-also-have-opposite-signs-find-the-set-of-possible-values-of-n-

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Question Number 207546 by pete last updated on 18/May/24
The real roots of the equation x^2 +6x+c=0 differ by 2n, where n is a real non−zero. Show that n^2 =9−c Given that the roots also have opposite signs, find the set of possible values of n
Therealrootsoftheequationx2+6x+c=0differby2n,wherenisarealnon−zero.Showthatn2=9−cGiventhattherootsalsohaveoppositesigns,findthesetofpossiblevaluesofn
Answered by A5T last updated on 18/May/24
x_1 +x_2 =−6; x_1 x_2 =c; ∣x_1 −x_2 ∣=2n ⇒4n^2 =(x_1 −x_2 )^2 =(x_1 +x_2 )^2 −4x_1 x_2 ⇒4n^2 =36−4c⇒n^2 =9−c x_(1,2) =((−6+_− (√(36−4c)))/2)=−3+_− (√(9−c))=−3+_− ∣n∣ when[−3+∣n∣≥0 and −3−∣n∣≤0 ⇒∣n∣≥3 and ∣n∣≥−3⇒∣n∣≥3⇒n≥3 or n≤−3..(i) when −3+∣n∣≤0 and −3−∣n∣≥0 ⇒∣n∣≤3 and ∣n∣≤−3⇒∅...(ii) ⇒n≥3 or n≤−3
x1+x2=−6;x1x2=c;∣x1−x2∣=2n⇒4n2=(x1−x2)2=(x1+x2)2−4x1x2⇒4n2=36−4c⇒n2=9−cx1,2=−6+−36−4c2=−3+−9−c=−3+−∣n∣when[−3+∣n∣⩾0and−3−∣n∣⩽0⇒∣n∣⩾3and∣n∣⩾−3⇒∣n∣⩾3⇒n⩾3orn⩽−3..(i)when−3+∣n∣⩽0and−3−∣n∣⩾0⇒∣n∣⩽3and∣n∣⩽−3⇒∅…(ii)⇒n⩾3orn⩽−3
Commented by pete last updated on 18/May/24
Thank you sir
Thankyousir

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