Question

It is known that a quadratic function

f(x)=ax^2+bx+2 (a<0)

satisfies the relation f(10)=0. Find the largest possible quantity of integer solutions of the inequality ax^4+bx^2+2>0.

Answer 1

$f(x) = a x\\^2 + bx + 2 \ (a<0)$. $a \ is \ negative.$ Therefore, the function can be written as: $f(x) = -a x\\^2 + bx + 2$

$f(10) = -a(10)\\^2 + b(10) + 2= 0$

$-100a + 10b + 2 = 0$

$10 b= 100a - 2$

$b= 10 a- \frac{2}{10}$

$ax\\^4 + bx\\^2 + 2>0$

$ax\\^4 + (10a - \frac{2}{10}) x\\^2 + 2 >0$

$f(10) = 0$

$a10\\^4 + (10a - \frac{2}{10}) 10\\^2 + 2 >0$

$10000a + (10a - \frac{2}{10}) 100 + 2 >0$

$10000a + 1000a -20 + 2 >0$

$11000a > 18$

$a>\frac{18}{11000}$

From the above the, largest possible quantity of integer is