Video: Solving Quadratic Equations by Factoring

Find the solution set of π‘₯Β² + 12π‘₯ = 0 in ℝ.


Video Transcript

Find the solution set of π‘₯ squared plus 12π‘₯ equals zero in the real numbers.

The solution set means we’re looking for the set of all values of our variable, which in this question is π‘₯, which satisfy the equation π‘₯ squared plus 12π‘₯ equals zero. And in case it’s relevant later, we’re only looking for values in the real numbers. Now the equation we’ve been given is a quadratic equation because the highest power of our variable π‘₯ is two. And in fact, it’s a relatively simple quadratic equation because there’s no constant term. We’re going to try to solve this quadratic equation by factoring. And looking at the two terms, we can see that they have a common factor of π‘₯.

We can therefore factor by π‘₯. Inside our parentheses, we need to work out what we have to multiply π‘₯ by to give each of the original two terms. To get π‘₯ squared, we have to multiply π‘₯ by π‘₯, and to get positive 12π‘₯, we have to multiply π‘₯ by positive 12. We therefore have our quadratic equation in its factored form: π‘₯ multiplied by π‘₯ plus 12 is equal to zero. And we can confirm that this is indeed equivalent by redistributing the parentheses if we wish.

Now, the key step to solving this quadratic equation once we found its factored form is to recall that if we have two factors multiplying to give zero, the only way this can happen is that at least one of the individual factors is itself equal to zero. Either π‘₯ equals zero or π‘₯ plus 12 is equal to zero. This gives two linear equations, which we can solve. Now the first π‘₯ equals zero actually needs no solution and the second π‘₯ plus 12 equals zero can be solved by simply subtracting 12 from each side to give π‘₯ equals negative 12. Therefore, there are two values in the solution set of this quadratic equation, the value zero and negative 12.

We can, of course, check each of these values by substituting them back into the original equation if we wish. For example, when π‘₯ equals negative 12, the expression on the left-hand side, π‘₯ squared plus 12π‘₯, becomes negative 12 squared plus 12 multiplied by negative 12. That’s 144 plus negative 144 or 144 minus 144 which is indeed equal to zero, the value on the right-hand side of the equation, confirming that negative 12 is indeed a correct value for π‘₯.

It’s also really important that at this stage here, we remember to factor our equation by π‘₯ and not simply divide by π‘₯. If we divide by π‘₯, we’re left with just π‘₯ plus 12 equals zero, which gives us only one solution π‘₯ equals negative 12. If instead we remember to factor our equation by π‘₯, then we also have the correct solution π‘₯ equals zero.

So, using the method of factoring, we’ve found the solution set of the given quadratic equation is the set of values zero, negative 12.

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