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if-a-b-198-what-is-the-largest-integer-root-which-the-equation-x-2-ax-b-0-may-have-

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Question Number 203367 by mr W last updated on 17/Jan/24
if a+b=198, what is the largest integer root which the equation x^2 +ax+b=0 may have?
ifa+b=198,whatisthelargestintegerrootwhichtheequationx2+ax+b=0mayhave?
Answered by ajfour last updated on 18/Jan/24
x^2 +ax−a=−(a+b)=−k (x−1)^2 +2(x−1)+a(x−1)=−k−1 (x−1+1+(a/2))^2 =−k−1+(1+(a/2))^2 x=−(a/2)±(√((1+(a/2))^2 −(k+1))) x−1=−(1+(a/2)){1+(√(1−(((k+1))/((1+(a/2))^2 )))) } If we let (1+(a/2))→ −∞ x−1=−2(1+(a/2)) x→∞
x2+ax−a=−(a+b)=−k(x−1)2+2(x−1)+a(x−1)=−k−1(x−1+1+a2)2=−k−1+(1+a2)2x=−a2±(1+a2)2−(k+1)x−1=−(1+a2){1+1−(k+1)(1+a2)2}Ifwelet(1+a2)→−∞x−1=−2(1+a2)x→∞
Answered by AST last updated on 26/Jan/24
−a=x_1 +x_2 ;b=x_1 x_2 198=a+b=x_1 (x_2 −1)−x_2 ⇒x_1 =((198+x_2 )/(x_2 −1))=1+((199)/(x_2 −1)) As x_2 →1^+ ;x_1 →∞⇒no largest integer root
−a=x1+x2;b=x1x2198=a+b=x1(x2−1)−x2⇒x1=198+x2x2−1=1+199x2−1Asx2→1+;x1→∞⇒nolargestintegerroot

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