Question
Find the discriminant.
6z^2 - 4z + 1 = 0
How many real solutions does the equation have?
6z^2 - 4z + 1 = 0
How many real solutions does the equation have?
Answers
GPT-4o mini
To find the discriminant of the quadratic equation \(6z^2 - 4z + 1 = 0\), we can use the formula for the discriminant, which is given by:
\[
D = b^2 - 4ac
\]
In this equation, the coefficients are:
- \(a = 6\)
- \(b = -4\)
- \(c = 1\)
Now, substituting the values of \(a\), \(b\), and \(c\) into the discriminant formula:
\[
D = (-4)^2 - 4 \cdot 6 \cdot 1
\]
\[
D = 16 - 24
\]
\[
D = -8
\]
The discriminant \(D\) is \(-8\).
Since the discriminant is negative (\(D < 0\)), this means that the quadratic equation \(6z^2 - 4z + 1 = 0\) has no real solutions. Instead, it has two complex solutions.
\[
D = b^2 - 4ac
\]
In this equation, the coefficients are:
- \(a = 6\)
- \(b = -4\)
- \(c = 1\)
Now, substituting the values of \(a\), \(b\), and \(c\) into the discriminant formula:
\[
D = (-4)^2 - 4 \cdot 6 \cdot 1
\]
\[
D = 16 - 24
\]
\[
D = -8
\]
The discriminant \(D\) is \(-8\).
Since the discriminant is negative (\(D < 0\)), this means that the quadratic equation \(6z^2 - 4z + 1 = 0\) has no real solutions. Instead, it has two complex solutions.