To determine if Trish found the correct zeros of the quadratic function [tex]\( f(x) = 2x^2 - 3x + 3 \)[/tex], we need to analyze the function and the steps provided.
First, the zeros of a quadratic function can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the quadratic function [tex]\( f(x) = 2x^2 - 3x + 3 \)[/tex] has the coefficients:
[tex]\[ a = 2, \quad b = -3, \quad c = 3 \][/tex]
Let's calculate the discriminant [tex]\( \Delta \)[/tex] of the quadratic function, where:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the given coefficients:
[tex]\[ \Delta = (-3)^2 - 4(2)(3) \][/tex]
[tex]\[ \Delta = 9 - 24 \][/tex]
[tex]\[ \Delta = -15 \][/tex]
The discriminant is [tex]\( -15 \)[/tex]. Since the discriminant is negative, this tells us that the quadratic function [tex]\( f(x) = 2x^2 - 3x + 3 \)[/tex] has no real number zeros. This means the quadratic equation has two complex conjugate roots.
Now, let's analyze Trish's steps:
1. [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]
2. [tex]\( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(3)}}{2(2)} \)[/tex]
3. [tex]\( x = \frac{3 \pm \sqrt{9 + 32}}{4} \)[/tex]
4. [tex]\( x = \frac{3 \pm \sqrt{41}}{4} \)[/tex]
Trish's steps indicate that she computed the expression inside the square root incorrectly. The correct discriminant is [tex]\( -15 \)[/tex], not [tex]\( 41 \)[/tex].
Since the discriminant is negative, there are no real number solutions to the quadratic equation. Therefore, Trish's conclusion that there are two real number zeros is incorrect.
So, the correct answer is:
[tex]\[ \boxed{\text{No, the function has no real number zeros.}} \][/tex]
