Question

suppose a, b are Integers and a+b is a root of x^2+ax+b=0. what is the maximum possible value of b^2?

Answer 1

Answer:

Maximum possible value of $\bold{\mathrm{b}^{2}}$ is 81.

Step-by-step explanation:

Step 1:

(a + b) 2 + a(a + b) + b = 0  

=> $b^2$ + (3a + 1)b + 2a2 = 0

Step 2:

Modify the above equation to find the value of b.

$\begin{array}{l}{b=\frac{-(3 a+1) \pm \sqrt{(3 a+1)^{2}-8^{2}}}{2}} \\ {b=\frac{-(3 a+1) \pm \sqrt{a^{2}+6 a+1}}{2}}\end{array}$

Step 3:

As a, b belongs to Integer, $a^{2}+6 a+1$ must be a perfect  a(a + 6) + 1 must be a perfect square  

Step 4:

Possible values of a are

Therefore a = 0 or a = –6 If a = 0, b = –1 or 0 If a = –6, b = 9 or 8.