What are the real roots of the equation 6x^4+x^3+52x^2+9x-18=0 if one root is 3i?

Asked on by clockbeige

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Since one root is a complex number, then the other root is the conjugate of the complex number.

x1 = 3i => x2 = -3i

The number of the roots of the equation is 4 and 2 of them are complex roots. Then, the other 2, could be the real roots of the equation.

Since x = 3i and x = -3i are the roots of the equation, the polynomial 6x^4+x^3+52x^2+9x-18 is divided by (x - 3i)(x + 3i).

We'll write the reminder theorem:

6x^4+x^3+52x^2+9x-18 = (x^2 + 9)(ax^2 + bx + c)

We'll remove the brackets:

6x^4+x^3+52x^2+9x-18 = ax^4 + bx^3 + cx^2 + 9ax^2 + 9bx + 9c

6x^4+x^3+52x^2+9x-18 = ax^4 + bx^3 + x^2(c + 9a) + 9bx + 9c

Comparing, we'll get:

a = 6

b = 1

9c = -18

c = -2

The quotient ax^2 + bx + c = 6x^2 + x - 2

If 6x^4+x^3+52x^2+9x-18 = 0, then 6x^2 + x - 2 = 0, too.

We'll apply quadratic formula:

x1 = [-1 + sqrt(1 + 48)]/12

x1 = (-1+7)/12

x1 = 1/2

x2 = -8/12

x2 = -2/3

The real roots of the equation are x = 1/2 and x = -2/3.

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