use the discriminant to solve for the smallest value of b that would make the roots of the equation x^2+bx+5=o imaginary
is this algebra or pre-algebra
to have imaginary roots
b^2 - 4ac < 0
b^2 - 4(1)(5) <0
b^ < 20
± b < √20
b < √20 AND b > -√20
so -√20 < b < √20
It would be that number "just to the right" of -√20
but of course we cannot name that number.
If you think you have found it, I can find a number which is closer to it.
Poorly worded question!
To solve for the smallest value of b that would make the roots of the equation x^2 + bx + 5 = 0 imaginary, we need to use the discriminant.
The discriminant is the expression found under the square root in the quadratic formula and is given by Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation.
In this case, the quadratic equation is x^2 + bx + 5 = 0. Comparing the equation with the standard form ax^2 + bx + c = 0, we can see that a = 1, b = b, and c = 5.
For the roots to be imaginary, the discriminant Δ must be negative. Therefore, we need to find the smallest value of b that makes Δ negative.
Δ = b^2 - 4ac
Δ = b^2 - 4(1)(5)
Δ = b^2 - 20
Since we want Δ to be negative, we can set Δ < 0 and solve for b:
b^2 - 20 < 0
b^2 < 20
Taking the square root of both sides:
√(b^2) < √20
|b| < √20
Considering that we are only interested in the smallest value of b, we take the negative square root:
-b < -√20
b > √20
Therefore, the smallest value of b that would make the roots of the equation x^2 + bx + 5 = 0 imaginary is b > √20.