To determine which quadratic equation has exactly one real number solution, we need to examine each equation to find its discriminant.
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is calculated using the formula:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
A quadratic equation will have:
- Two distinct real number solutions if [tex]\(\Delta > 0\)[/tex]
- Exactly one real number solution if [tex]\(\Delta = 0\)[/tex]
- No real number solutions if [tex]\(\Delta < 0\)[/tex]
Let's calculate the discriminant for each equation:
1. For the equation [tex]\(0 = 2x^2 - 4x + 1\)[/tex]:
[tex]\[
a = 2, \quad b = -4, \quad c = 1
\][/tex]
[tex]\[
\Delta = (-4)^2 - 4 \cdot 2 \cdot 1 = 16 - 8 = 8
\][/tex]
Since [tex]\(\Delta = 8\)[/tex], which is greater than 0, this equation has two distinct real number solutions.
2. For the equation [tex]\(0 = 2x^2 - 5x + 3\)[/tex]:
[tex]\[
a = 2, \quad b = -5, \quad c = 3
\][/tex]
[tex]\[
\Delta = (-5)^2 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1
\][/tex]
Since [tex]\(\Delta = 1\)[/tex], which is greater than 0, this equation has two distinct real number solutions.
3. For the equation [tex]\(0 = -2x^2 - 4x - 2\)[/tex]:
[tex]\[
a = -2, \quad b = -4, \quad c = -2
\][/tex]
[tex]\[
\Delta = (-4)^2 - 4 \cdot (-2) \cdot (-2) = 16 - 16 = 0
\][/tex]
Since [tex]\(\Delta = 0\)[/tex], this equation has exactly one real number solution.
4. For the equation [tex]\(0 = -2x^2 - 3x - 1\)[/tex]:
[tex]\[
a = -2, \quad b = -3, \quad c = -1
\][/tex]
[tex]\[
\Delta = (-3)^2 - 4 \cdot (-2) \cdot (-1) = 9 - 8 = 1
\][/tex]
Since [tex]\(\Delta = 1\)[/tex], which is greater than 0, this equation has two distinct real number solutions.
Among the given equations, the one that has exactly one real number solution is:
\[
0 = -2x^2 - 4x - 2
