The roots of the equation z^3 + 2z^2 +3z - 4 = 0, are a, b and c . Show that a^2 + b^2 +c^2 = -2

If the roots of this cubic equation are a, b and c, then the equation can be written

(z - a)(z - b)(z - c) = 0

multiplying this out gives:

z^3 - az^2 - bz^2 - cz^2 + abz + acz + bcz - abc

grouping terms with z to the same power in them, this becomes:

z^3  + (-a  - b - c)z^2 + (ab + ac + bc)z - abc

Equating the coefficients in the equation above with the coefficients of the equation in the question gives the following equations:

(-a -b - c) =2 


a + b + c = -2


(ab + ac + bc) = 3


-abc = -4.

To show that a^2 + b^2 + c^2 = -2, first we need to manipulate our expressions above to get an expression with a^2 + b^2 +c^2 in it. The most obvious way to do this is by squaring (a + b +c) i.e.

(a + b + c)^2 = (-2)^2

which multiplies out to give

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 4

-> a^2 + b^2 + c^2 + 2(ab + ac + bc) = 4

We now that (ab + ac + bc) = 3, so substitue this into the expression above

a^2 + b^2 + c^2 + 2(3) = 4

Finally, take away 6 from both sides to get,

a^2 + b^2 + c^2 = -2

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