Find the greatest integer value of b for which the expression 9x^3+4x^2+11x+7/x^2+bx+8 has a domain for all real numbers. Thank you!
The greatest integer is 5.
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To find the greatest integer value of b for which the expression 9x^3 + 4x^2 + 11x + 7 / x^2 + bx + 8 has a domain for all real numbers, we need to ensure that the denominator x^2 + bx + 8 does not have any real roots.
For a quadratic equation ax^2 + bx + c, the discriminant Δ is calculated as Δ = b^2 - 4ac.
If the discriminant is greater than zero (Δ > 0), the quadratic equation has two distinct real roots.
If the discriminant is equal to zero (Δ = 0), the quadratic equation has one real root (the roots are coincident).
If the discriminant is less than zero (Δ < 0), the quadratic equation has no real roots.
In this case, we want to find the value of b for which the discriminant is less than zero, so that the quadratic equation x^2 + bx + 8 has no real roots.
Setting the discriminant Δ < 0, we have:
b^2 - 4ac < 0
Substituting the values a = 1, b = b, and c = 8, we get:
b^2 - 4(1)(8) < 0
b^2 - 32 < 0
Rearranging the inequality:
b^2 < 32
Taking the square root of both sides, we have:
|b| < √32
√32 ≈ 5.6568
Since we are seeking the greatest integer value of b, we round down to the nearest whole number:
|b| < 5
Therefore, the greatest integer value of b for which the expression has a domain for all real numbers is b = 4.
To determine the domain of the given expression, we need to find the values of x for which the denominator is not equal to zero. In other words, we are looking for the values of x that make the expression well-defined.
In this case, the denominator is x^2 + bx + 8. To ensure that it is not equal to zero for any real numbers, we need to find the values of b that make the quadratic equation have no real solutions.
The quadratic equation x^2 + bx + 8 = 0 has real solutions if its discriminant (b^2 - 4ac) is greater than or equal to zero, where a = 1, b = b, and c = 8.
So, we have: b^2 - 4ac ≥ 0
Substituting the values: b^2 - 4(1)(8) ≥ 0
Simplifying: b^2 - 32 ≥ 0
To find the greatest integer value of b that satisfies this inequality, we need to find the largest value of b for which b^2 - 32 ≥ 0.
Taking the square root of both sides of the inequality, we have:
√(b^2 - 32) ≥ √0
b^2 - 32 ≥ 0
Rearranging the equation, we have:
b^2 ≥ 32
Taking the square root of both sides, we have:
|b| ≥ √32
|b| ≥ 4√2
Since we are looking for the greatest integer value of b, we need to round up the value of 4√2 to the nearest whole number. Since √2 is approximately 1.414, the greatest integer value of b would be 4.
So, the greatest integer value of b for which the expression 9x^3 + 4x^2 + 11x + 7/x^2 + bx + 8 has a domain for all real numbers is 4.
the denominator cannot be zero. So,
x^2+bx+8 must have no real roots. For that to happen, we need the discriminant to be negative:
b^2-32 < 0
So |b| <= √32
5 is the greatest integer less than √32