Let's start by solving the given quadratic equation [tex]\(2x^2 - 4x = -2\)[/tex]. First, we need to rewrite it in standard form, which is [tex]\(ax^2 + bx + c = 0\)[/tex].
Rearranging the equation, we get:
[tex]\[
2x^2 - 4x + 2 = 0
\][/tex]
Here, the coefficients are [tex]\(a = 2\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 2\)[/tex].
To determine the number and type of solutions for this quadratic equation, we use the discriminant (D), which is given by the formula:
[tex]\[
D = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the formula:
[tex]\[
D = (-4)^2 - 4 \cdot 2 \cdot 2
\][/tex]
[tex]\[
D = 16 - 16
\][/tex]
[tex]\[
D = 0
\][/tex]
The discriminant [tex]\(D\)[/tex] is 0.
The nature of the solutions of a quadratic equation can be determined based on the value of the discriminant:
- If [tex]\(D > 0\)[/tex], there are two distinct real solutions.
- If [tex]\(D = 0\)[/tex], there is exactly one real solution.
- If [tex]\(D < 0\)[/tex], there are two complex (nonreal) solutions.
Since the discriminant [tex]\(D = 0\)[/tex] in this case, the quadratic equation [tex]\(2x^2 - 4x + 2 = 0\)[/tex] has exactly one real solution.
Therefore, the answer is:
- one real solution
