Question Video: Finding the Solution Set of a Quadratic Equation Mathematics

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


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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.