Solve the equation and write complex solutions in standard form.

Question

Quizplus · Accepted Answer

Since the equation has a variable raised to an odd power, it is likely that there will be complex solutions. To solve, begin by simplifying the expression in the parentheses:

$(x^2+16x+65)^3 = (x^2+16x+65)(x^2+16x+65)(x^2+16x+65)$

Expanding this out and collecting like terms gives:

$x^6 + 48x^5 + 820x^4 + 7040x^3 + 32625x^2 + 81120x + 274625 = 0$

Now substitute $y = x^2+16x+65$. This gives:

$y^3 = -(4x+65)^3$

$y^3 + (4x+65)^3 = 0$

Using the sum of cubes factorization, we get:

$(y + 4x + 65)(y^2 - 4xy + 16x^2 + 130y + 260x + 4225) = 0$

The first factor gives $y = -4x-65$, which can be substituted back into $y^2 - 4xy + 16x^2 + 130y + 260x + 4225 = 0$ to obtain a quadratic equation in $x$:

$13x^2 + 52x + 41 = 0$

The solutions of this quadratic equation are $x = \frac{-26 \pm i\sqrt{117}}{13}$. Substituting these back into $y = -4x-65$ gives the complex solutions in standard form:

$y = -4\left(\frac{-26 \pm i\sqrt{117}}{13}\right) - 65$

$y = \frac{3}{13} \pm \frac{4\sqrt{117}}{13}i$

Finally, substituting $y = x^2+16x+65$ back into these solutions gives the complex solutions to the original equation:

$x^2+16x+65 = \frac{3}{13} + \frac{4\sqrt{117}}{13}i$ or $x^2+16x+65 = \frac{3}{13} - \frac{4\sqrt{117}}{13}i$

Solve the Equation and Write Complex Solutions in Standard Form