Question Video: Solving Quadratic Equations by Factoring | Nagwa Question Video: Solving Quadratic Equations by Factoring | Nagwa

# Question Video: Solving Quadratic Equations by Factoring Mathematics • First Year of Secondary School

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Find the solution set of (𝑥 + 2)/2 = 12/𝑥 in ℝ.

04:13

### Video Transcript

Find the solution set of 𝑥 plus two over two equals 12 over 𝑥 in the real numbers.

The solution set means we’re looking for the set of all values of the variable — in this case, that’s 𝑥 — which satisfy the given equation. And we’re told that we’re only interested in values of 𝑥 which exist in the real numbers.

Now, at first glance, this equation looks a little complex because it involves fractions. And in fact we can see that the variable we’re solving for, 𝑥, is in the denominator of one of these fractions. However, both denominators can be eliminated by cross multiplying. Multiplying both sides by two and 𝑥, we achieve 𝑥 multiplied by 𝑥 plus two is equal to 12 multiplied by two. On the right-hand side, 12 multiplied by two is 24. And on the left-hand side, we need to distribute this factor of 𝑥 over our parentheses. 𝑥 multiplied by 𝑥 gives 𝑥 squared, and 𝑥 multiplied by two gives two 𝑥. So we have 𝑥 squared plus two 𝑥 is equal to 24.

Next, we wish to collect all of the terms on the same side of the equation. And subtracting 24 from each side will achieve this, giving 𝑥 squared plus two 𝑥 minus 24 is equal to zero. So we’ve taken a relatively complicated-looking equation, which we may not immediately recognize as a quadratic. And by rearranging, we’ve written it in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero, the most easily recognizable form of a quadratic equation.

We’re going to try to solve this equation by factoring. First, we note that the coefficient of 𝑥 squared is one, which means that the first term in each of our set of parentheses is simply 𝑥 because 𝑥 multiplied by 𝑥 gives 𝑥 squared. To complete our parentheses, we’re then looking for two numbers which sum to the coefficient of 𝑥 — that’s positive two — and whose product is the constant term — that’s negative 24.

Listing the factor pairs of 24 can help here. The factor pairs are one and 24, two and 12, three and eight, and four and six. We know though that in order to make negative 24, we need one of these numbers to be negative and the other to be positive. If we choose the final factor pair and if we make the four negative, so we have negative four and six, then these two numbers do have a product of negative 24. And they also have a sum of positive two. So these are the two numbers we’re looking for to complete our parentheses. The factored form of our quadratic then is 𝑥 plus six multiplied by 𝑥 minus four is equal to zero.

Remember, we can write these factors in any order. To solve our quadratic equation from this stage, we recall that if two factors multiply to give zero, this is only possible if at least one factor is itself zero. So we take each factor in turn, set it equal to zero, and then solve the resulting linear equation. We have either 𝑥 plus six equals zero or 𝑥 minus four equals zero.

The first equation can be solved by subtracting six from each side to give 𝑥 equals negative six and the second by adding four to each side to give 𝑥 equals four. We found then that there are two values in the solution set of this equation, the values negative six and four. We can of course check either or both of these values by substituting back into the original equation. For example, when 𝑥 equals four, the left-hand side becomes four plus two over two. That’s six over two, which is equal to three. And the right-hand side becomes 12 over four, which is also equal to three. And so this confirms that 𝑥 equals four is a valid solution for this equation.

We can then do the same for 𝑥 equals negative six if we wish. So we’ve found the solution set of this equation. It’s the set of values negative six, four.

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