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

$i f a + b = 198, w h a t i s t h e l a r g e s t$ $i n t e g e r r o o t w h i c h t h e e q u a t i o n$ $x^{2} + a x + b = 0 m a y h a v e ?$
	Answered by ajfour last updated on 18/Jan/24
	$x^2 +ax−a=−(a+b)=−k (x−1)^2 +2(x−1)+a(x−1)=−k−1 (x−1+1+(a/2))^2 =−k−1+(1+(a/2))^2 x=−(a/2)±(√((1+(a/2))^2 −(k+1))) x−1=−(1+(a/2)){1+(√(1−(((k+1))/((1+(a/2))^2 )))) } If we let (1+(a/2))→ −∞ x−1=−2(1+(a/2)) x→∞$
	$x^{2} + a x - a = - (a + b) = - k$ ${(x - 1)}^{2} + 2 (x - 1) + a (x - 1) = - k - 1$ ${(x - 1 + 1 + \frac{a}{2})}^{2} = - k - 1 + {(1 + \frac{a}{2})}^{2}$ $x = - \frac{a}{2} \pm \sqrt{{(1 + \frac{a}{2})}^{2} - (k + 1)}$ $x - 1 = - (1 + \frac{a}{2}) {1 + \sqrt{1 - \frac{(k + 1)}{{(1 + \frac{a}{2})}^{2}}}}$ $I f w e l e t (1 + \frac{a}{2}) \to - \infty$ $x - 1 = - 2 (1 + \frac{a}{2})$ $x \to \infty$
	Answered by deleteduser1 last updated on 26/Jan/24

	$- a = x_{1} + x_{2}; b = x_{1} x_{2}$ $198 = a + b = x_{1} (x_{2} - 1) - x_{2} \Rightarrow x_{1} = \frac{198 + x_{2}}{x_{2} - 1} = 1 + \frac{199}{x_{2} - 1}$ $A s x_{2} \to 1^{+}; x_{1} \to \infty \Rightarrow n o l a r g e s t i n t e g e r r o o t$
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