Video: Finding the Solution Set of a Quadratic Equation

Find the solution set of 2π‘₯Β³ = 32π‘₯ in ℝ.

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Video Transcript

Find the solution set of two π‘₯ cubed equals 32π‘₯ in the real numbers.

First, we should get our equation equal to zero. And we want to keep the variable with the highest exponent positive, so π‘₯ cubed β€” that needs to stay positive. So the two π‘₯ cubed will stay positive. So we need to subtract the 32π‘₯ from both sides of the equation, which leaves us with two π‘₯ cubed minus 32π‘₯ equals zero.

So since there are two terms, we can factor by first taking out a greatest common factor. And we can take out a two and an π‘₯. If we take out two π‘₯ from two π‘₯ cubed, we’re left with π‘₯ squared. If we take out two π‘₯ from 32π‘₯, we’re left with 16.

Now, inside the parentheses, we have a difference of two squares because π‘₯ squared we can square root and 16 we can square root. The formula for difference of squares is π‘Ž squared minus 𝑏 squared equals π‘Ž plus 𝑏 times π‘Ž minus 𝑏. So if we have π‘₯ squared minus 16, the square root of π‘₯ squared is π‘₯ and the square root of 16 is four. So we will have π‘₯ plus four and π‘₯ minus four.

So now, we’ve completely factored our equation. So to find the solution, we need to take each factor and set it equal to zero. So we set two π‘₯ equal to zero, we set π‘₯ plus four equal to zero, and we set π‘₯ minus four equal to zero.

So for two π‘₯ equals zero, we need to divide both sides of the equation by two. And we get that π‘₯ is equal to zero. For π‘₯ plus four equals zero, we need to subtract four from both sides of the equation. And we’re left with π‘₯ equals negative four. For π‘₯ minus four equals zero, we need to add four to both sides of the equation. And we have π‘₯ equals four.

So our solution set would be zero, four, and negative four. And the order does not matter.

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