Question Video: Solving Quadratic Equations by Factorisation Mathematics

Find the solution set of π‘₯(π‘₯ βˆ’ 19) = βˆ’15π‘₯ in ℝ.

03:33

Video Transcript

Find the solution set of π‘₯ multiplied by π‘₯ minus 19 equals negative 15π‘₯ in the set of real numbers.

The solution set means we’re looking to find all the values of the variable, in this case π‘₯, which satisfy the given equation. We were also told in the question that we’re only interested in values of π‘₯ which are real numbers. Let’s begin by distributing the parentheses on the left-hand side of our equation. π‘₯ multiplied by π‘₯ gives π‘₯ squared, and π‘₯ multiplied by negative 19 gives negative 19π‘₯. So our equation becomes π‘₯ squared minus 19π‘₯ equals negative 15π‘₯.

Next, we want to collect all the terms on the same side of the equation, which we can do by adding 15π‘₯ to each side. On the left-hand side, we have π‘₯ squared minus four π‘₯. And on the right-hand side, we now have zero. What we should now see, which may not have been obvious before, is that we have a quadratic equation. And in fact, it’s one of the simpler types of quadratic equation because we have no constant term or our constant term is just zero. We can solve this quadratic equation by factoring.

Our two terms of π‘₯ squared and negative four π‘₯ share a common factor of π‘₯. So we can factor by π‘₯. Inside the parentheses, we need the two terms that we have to multiply π‘₯ by to give the original expression on the left-hand side. π‘₯ multiplied by π‘₯ gives π‘₯ squared, and π‘₯ multiplied by negative four gives negative four π‘₯. So the factored form of our quadratic is π‘₯ multiplied by π‘₯ minus four is equal to zero. Now we have two factors multiplying to give zero. And the only way this can happen is if at least one of the individual factors is itself equal to zero.

So to find all the values in the solution set of this equation, we take each factor in turn, set it equal to zero, and then solve the resulting equation. We have either π‘₯ equals zero, which requires no further work, or π‘₯ minus four equals zero. This second equation can be solved by simply adding four to each side, giving π‘₯ equals four. Therefore, there are two values in the solution set of this equation, the values zero and four.

We can of course check each of these values by substituting into each side of the original equation. For example, when π‘₯ equals four, π‘₯ multiplied by π‘₯ minus 19 gives four multiplied by negative 15, which is negative 60. And on the right-hand side, negative 15π‘₯ is negative 15 multiplied by four, which is also negative 60. And so this confirms that π‘₯ equals four is a valid solution to this equation. We do need to be especially careful when factoring this quadratic. A common mistake is to simply divide the quadratic π‘₯ squared minus four π‘₯ equals zero by π‘₯, leaving π‘₯ minus four equals zero.

However, if we do this, we lose one of the values in our solution set. Our only answer would be π‘₯ equals four. And we would’ve lost the valid solution π‘₯ equals zero. So by factoring this quadratic equation once we have one side equal to zero and then solving the resulting two linear equations, we found the solution set of this equation. It is the set of values zero, four.

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