Question

what are all the solutions to the equation (10x^4 + 30x^3 + 35x^2 = 0)?

(0, \frac{3 + i\sqrt{5}}{2}, \frac{3 - i\sqrt{5}}{2})

(0, \frac{-3 + i\sqrt{5}}{2}, \frac{-3 - i\sqrt{5}}{2})

(0, 6 + 2i\sqrt{5}, 6 - 2i\sqrt{5})

(0, -6 + 2i\sqrt{5}, -6 - 2i\sqrt{5})

Explanation:

Factor out the greatest common factor

$$ 10x^4 + 30x^3 + 35x^2 = 5x^2(2x^2 + 6x + 7) = 0 $$

This yields the first solution:

$$ 5x^2 = 0 \implies x = 0 $$

Solve the remaining quadratic equation

$$ 2x^2 + 6x + 7 = 0 $$

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

$$ x = \frac{-6 \pm \sqrt{6^2 - 4(2)(7)}}{2(2)} = \frac{-6 \pm \sqrt{36 - 56}}{4} = \frac{-6 \pm \sqrt{-20}}{4} $$

Simplify the complex roots

$$ x = \frac{-6 \pm 2i\sqrt{5}}{4} = \frac{-3 \pm i\sqrt{5}}{2} $$

Thus, the complete set of solutions is:

$$ 0, \frac{-3 + i\sqrt{5}}{2}, \frac{-3 - i\sqrt{5}}{2} $$

Answer:

• (A) \(0, \frac{3+i\sqrt{5}}{2}, \frac{3-i\sqrt{5}}{2}\)
• (B) \(0, \frac{-3+i\sqrt{5}}{2}, \frac{-3-i\sqrt{5}}{2}\) (Correct answer)
• (C) \(0, 6+2i\sqrt{5}, 6-2i\sqrt{5}\)
• (D) \(0, -6+2i\sqrt{5}, -6-2i\sqrt{5}\)