Question

Trish is solving for the zeros of the quadratic function f(x) = 2x2 – 3x + 3. x = StartFraction negative b plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction x = StartFraction negative (negative 3) plus or minus StartRoot (negative 3) squared minus 4 (2)(3) EndRoot Over 2 (2) EndFraction x = StartFraction 3 plus or minus StartRoot b 9 + 32 EndRoot Over 4 EndFraction x = StartFraction 3 plus or minus StartRoot 41 EndRoot Over 4 EndFraction Did Trish find the correct zeros of this function? Explain. Yes, those are the two real number zeros. No, the two real number zeros are StartFraction negative 3 plus or minus StartRoot 41 EndRoot Over 4 EndFraction. No, the two real number zeros are StartFraction 3 plus or minus StartRoot 41 EndRoot Over negative 4 EndFraction. No, the function has no real number zeros.

Answer 1

Answer:

On online classes the answer is: NO, THE FUNCTION HAS NO REAL NUMBER ZEROS

Step-by-step explanation:

its D on online work

Answer 2

The zeroes of the quadratic function  f(x) = 2x² – 3x + 3 is imaginary.

What is a quadratic equation?

The quadratic equation is given as ax² + bx + c = 0. Then the degree of the equation will be 2. Then we have

Trish is solving for the zeros of the quadratic function f(x) = 2x² – 3x + 3.

Then the zeroes of the quadratic equation are given by the formula method. Then we have

$\rm x = \dfrac{- (-3) \pm \sqrt{(-3)^2 - 4 \times 2 \times 3}}{2 \times 2}\\\\x = \dfrac{3 \pm \sqrt{-15}}{4}\\\\x = \dfrac{3 \pm \sqrt{15} \iota }{4}$

Then the zeroes of the quadratic function  f(x) = 2x² – 3x + 3 is imaginary.

More about the quadratic equation link is given below.

https://brainly.com/question/2263981

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