Video: Finding the Solution Set of a Quadratic Equation

Video Transcript

Find the solution set of two 𝑥 cubed equals 32𝑥 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 𝑥 cubed — that needs to stay positive. So the two 𝑥 cubed will stay positive. So we need to subtract the 32𝑥 from both sides of the equation, which leaves us with two 𝑥 cubed minus 32𝑥 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 𝑥. If we take out two 𝑥 from two 𝑥 cubed, we’re left with 𝑥 squared. If we take out two 𝑥 from 32𝑥, we’re left with 16.

Now, inside the parentheses, we have a difference of two squares because 𝑥 squared we can square root and 16 we can square root. The formula for difference of squares is 𝑎 squared minus 𝑏 squared equals 𝑎 plus 𝑏 times 𝑎 minus 𝑏. So if we have 𝑥 squared minus 16, the square root of 𝑥 squared is 𝑥 and the square root of 16 is four. So we will have 𝑥 plus four and 𝑥 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 𝑥 equal to zero, we set 𝑥 plus four equal to zero, and we set 𝑥 minus four equal to zero.

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

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