Answer:
1 + i[tex]\sqrt{5}[/tex], 1 - i[tex]\sqrt{5}[/tex]
Step-by-step explanation:
A quadratic function is f(x) = a[tex]x^{2}[/tex] + bx + c
Your function with the given values is f(x) = 2[tex]x^{2}[/tex] - 4x + 12
To solve a quadratic function, you can use the quadratic formula
(-b +/- [tex]\sqrt{b^{2} - 4ac}[/tex]) / 2a
Fill in with your given values.
(4 +/- [tex]\sqrt{(-4^{2} - 4 (2) (12)}[/tex] ) / 2(2)
Simplify
(4 +/- [tex]\sqrt{16 - 96}[/tex]) / 4
(4 +/- [tex]\sqrt{-80}[/tex] ) / 4
Because there is a negative number under the square root making it an imaginary number put an i by it.
(4 +/- i[tex]\sqrt{80}[/tex]) / 4
80 is divisible by 16, a perfect square. 80 / 16 = 5
(4 +/- 4i[tex]\sqrt{5}[/tex]) / 4
Divide by 4
1 +/- i[tex]\sqrt{5}[/tex]
Your teacher may want you to list your radical without the i and just as [tex]\sqrt{-5}[/tex], so keep that in mind.
Don't forget to list it both as addition and subtraction.
