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If roots of the equation $x^{2} + ax + b = 0$ are positive integers and $a + b = 198$ then values of $a$ and $b$ are respectively.

My solution: Positive real roots means $D > 0$. Therefore $b^{2}-4ac>0$. On putting value of $b$ from $a+b$ gives $a^{2} +4a -792=0$. I could not proceed beyong this. I also tried using sum/product of roots but I did not know what to do with it or the fact that the roots are integral.

Kenta S
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Shaurya Goyal
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1 Answers1

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$x^2 + ax + b = 0$

say, it has integer roots $m$ and $n$.

So, $(x-m)(x-n) = 0$

Since, $a+b = 198$, $mn-(m+n) = 198$

or, $(m-1)(n-1) = 199$

As 199 is a prime number, $m-1 = 1, n-1 = 199$ or vice versa.

So, roots are $2$ and $200$ and $a = -202, b = 400$.

Math Lover
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