Question - Solving a Cubic Equation by Factoring and Quadratic Formula
Solution:
To solve the equation $$18x^3 + 12x^2 + 2x = 0$$, you can start by factoring out the common factor, which is $$2x$$:$$2x(9x^2 + 6x + 1) = 0$$Now, you have a product of two expressions that equals zero. According to the zero-product property, one or both of these expressions must equal zero for the equation to be true. We can set each expression equal to zero and solve for $$x$$:1. $$2x = 0$$ gives us one solution:$$x = 0$$2. $$9x^2 + 6x + 1 = 0$$ is a quadratic equation. It doesn't factor nicely, so we can use the quadratic formula to find the solutions:$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$where $$a = 9$$, $$b = 6$$, and $$c = 1$$.Plugging in these values, we get:$$x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9}$$$$x = \frac{-6 \pm \sqrt{36 - 36}}{18}$$$$x = \frac{-6 \pm \sqrt{0}}{18}$$$$x = \frac{-6}{18}$$The discriminant ($$b^2 - 4ac$$) being zero indicates that there is only one real solution for this part of the equation, which is:$$x = \frac{-1}{3}$$So, the equation has the following solutions:$$x = 0$$ and $$x = -\frac{1}{3}$$
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