if the discriminant of (5u^2)+bu+2=0 is 41, what are all the values of b
what is the final answer?
The discriminant of a quadratic equation
Ax²+Bx+C=0
is
Δ=B²-4AC
Here A=5,B=b,C=2
so
Δ=b²-4*5*2=41
Solve for b to get b=sqrt(1)
To find the values of \(b\) for which the discriminant of the quadratic equation \(5u^2 + bu + 2 = 0\) is 41, we need to use the formula for the discriminant.
The discriminant (\(\Delta\)) of a quadratic equation in the form \(au^2 + bu + c = 0\) is calculated using the formula: \(\Delta = b^2 - 4ac\).
We are given that the discriminant (\(\Delta\)) is 41, so we can set up the equation:
\(41 = b^2 - 4(5)(2)\)
Simplifying the equation:
\(41 = b^2 - 40\)
Now, let's bring 40 to the other side of the equation:
\(b^2 = 81\)
To get the values of \(b\), we take the square root of both sides:
\(b = \pm\sqrt{81}\)
Simplifying further, we have:
\(b = \pm 9\)
Therefore, the values of \(b\) for which the discriminant is 41 are \(b = -9\) and \(b = 9\).