Find the greatest integer value of b for which the expression 9x^2+4x^2+11x+7/x^2+bx+8 has a domain for all real numbers? Thank you!
if the domain is all real numbers then the denominator can never be zero. That means it has no real roots. Thus, the discriminant is negative:
b^2-32 < 0
So, if 0<=b<=5 x^2+bx+8 is never zero
To find the greatest integer value of b for which the expression has a domain for all real numbers, we need to determine the values of b that would cause the denominator, x^2 + bx + 8, to be non-zero for all real numbers.
The expression 9x^2 + 4x^2 + 11x + 7 / (x^2 + bx + 8) is defined for all real numbers as long as x^2 + bx + 8 is not equal to zero. In other words, we need to find values of b such that the quadratic equation x^2 + bx + 8 = 0 does not have any real solutions.
For a quadratic equation to have no real solutions, its discriminant (b^2 - 4ac) must be negative. In this case, the discriminant is b^2 - 4(1)(8) = b^2 - 32.
So, we want the discriminant, b^2 - 32, to be negative.
b^2 - 32 < 0
b^2 < 32
|b| < √32
Now, let's find the square root of 32:
√32 = 4√2
So, |b| < 4√2.
Since we are looking for the greatest integer value of b, we need to find the largest integer that is less than 4√2. The largest integer less than 4√2 is 5.
Therefore, the greatest integer value of b for which the expression has a domain for all real numbers is b = 5.