Respuesta :

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For this case we have the following polynomial:

[tex]3x^2 - 3x + 1[/tex]

Applying the formula of the resolvent we have:

[tex]x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}[/tex]

Substituting values we have:

[tex]x=\frac{-(-3)+/-\sqrt{(-3)^2-4(3)(1)}}{2(3)}[/tex]

Rewriting we have:

[tex]x=\frac{3+/-\sqrt{9-12}}{6}[/tex]

[tex]x=\frac{3+/-\sqrt{-3}}{6}[/tex]

[tex]x=\frac{3+/-i\sqrt{3}}{6}[/tex]

Therefore, the roots of the polynomial are given by:

[tex]x=\frac{3+i\sqrt{3}}{6}[/tex]

[tex]x=\frac{3-i\sqrt{3}}{6}[/tex]

Answer:

The values of x that are roots of the polynomial are:

[tex]x=\frac{3+i\sqrt{3}}{6}[/tex]

[tex]x=\frac{3-i\sqrt{3}}{6}[/tex]

The roots of a polynomial are where the polynomial equals zero


3x² − 3x + 1 = 0

∴ 3x² − 3x = -1

∴ x² − x = -1/3

∴ x² − x + (-1/2)² = (-1/2)² − 1/3

given x² + bx + (b/2)² = (x + b/2)²

∴ (x − 1/2)² = 1/4 − 1/3

∴ (x − 1/2)² = 3/12 − 4/12

∴ (x − 1/2)² = -1/12

∴ x − 1/2 = ±√(-1/12)

This tells us there are no real roots and if you need real number solutions we stop here

∴ x − 1/2 = ±i/(2√3)

∴ x − 1/2 = ±i√(3)/6

∴ x = 1/2 ± i√(3)/6

∴ x = 3/6 ± i√(3)/6

∴ x = (3 − i√3)/6 and (3 + i√3)/6 <= the 2 complex roots


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