Question

Trish is solving for the zeros of the quadratic function f(x) = 2x2 – 3x + 3. x = x = x = x = 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 . No, the two real number zeros are . No, the function has no real number zeros.

Answer 1

Answer:

No, the function has no real number zeros

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$f(x)=2x^{2} -3x+3$  

equate the function to zero

so

$2x^{2}-3x+3=0$  

$a=2\\b=-3\\c=3$

substitute in the formula

$x=\frac{3(+/-)\sqrt{(-3)^{2}-4(2)(3)}} {2(2)}$

$x=\frac{3(+/-)\sqrt{9-24}} {4}$

$x=\frac{3(+/-)\sqrt{-15}} {4}$

remember that

$i=\sqrt{-1}$

$x=\frac{3+\sqrt{15}i} {4}$

$x=\frac{3-\sqrt{15}i} {4}$

The function has no real number zeros

Answer 2

2x²-3x+3=0
The function has no real zeroes. If the equation is in the quadratic form ax²+bx+c, then b²-4ac has to be greater than zero (using the quadratic formula). In this case 4-24<0
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