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Olympiad workbook
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Olympiad workbook
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IOQM
Chapter
Quadratic Equation

Olympiad workbook · IOQM · Quadratic Equation

Suppose a, b are integers and a+b is a root of x^2+ax+b=0 . What is...

Suppose a,b\mathrm{a}, \mathrm{b} are integers and a+b\mathrm{a}+\mathrm{b} is a root of x2+ax+b=0\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0. What is the maximum possible values of b2b^2 ?

Correct answer

81

Step-by-step solution

If " a+ba+b " is a root it satisfies the equation Hence (a+b)2+a(a+b)+b=0(a+b)^2+a(a+b)+b=0 ⇒2a2+3ba+\Rightarrow 2 \mathrm{a}^2+3 \mathrm{ba}+\left(\mathrm{b}^2+\mathrm{b}\right)=0Nowsince"a"isanintegerDiscriminantisaperfectsquareNow since " a " is an integer Discriminant is a perfect square⇒9 b2−8( b2+b)=p2 for same p∈Z⇒( b−4)2−16=p2⇒( b−4+b)(b−4−p)=16 b−4+p=±8, b−4−p=±2, b−4+p=b−4−p=±4\begin{aligned} & \Rightarrow 9 \mathrm{~b}^2-8\left(\mathrm{~b}^2+\mathrm{b}\right)=\mathrm{p}^2 \text { for same } \mathrm{p} \in \mathrm{Z} \\ & \Rightarrow (\mathrm{~b}-4)^2-16=\mathrm{p}^2 \\ & \Rightarrow (\mathrm{~b}-4+\mathrm{b})(\mathrm{b}-4-\mathrm{p})=16 \\ & \mathrm{~b}-4+\mathrm{p}= \pm 8, \mathrm{~b}-4-\mathrm{p}= \pm 2, \mathrm{~b}-4+\mathrm{p}=\mathrm{b}-4-\mathrm{p}= \pm 4 \end{aligned}So,So,b-4=5,-5,4,-4 \Rightarrow \mathrm{b}=9,-1,8,0 \Rightarrow \left(\mathrm{~b}^2\right)_{\max }=81
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