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

Find the solution set of (𝑥 + 2)(𝑥 − 7) = −8 in ℝ.

04:20

Video Transcript

Find the solution set of 𝑥 plus two multiplied 𝑥 minus seven equals negative eight in the real numbers.

The solution set means we’re looking to find the set of all values of the variable, in this case, that’s 𝑥, which satisfy the given equation. And we’re also told that we’re only looking for solutions in the real numbers. Now, at first glance, this quadratic may look as if it’s already in its factored form. But if we look carefully, we see that the right-hand side is equal to negative eight, not zero. In order to solve a quadratic equation using the method of factoring, we must make sure that one side of the equation is equal to zero. So, instead, what we’re going to do is distribute the parentheses on the left-hand side and then rearrange the resulting equation so that we have zero on one side.

We can distribute the parentheses using whichever method we wish. Perhaps the FOIL method is most efficient. Multiplying the first terms, 𝑥 multiplied by 𝑥 gives 𝑥 squared. Multiplying the outer terms, 𝑥 multiplied by negative seven gives negative seven 𝑥. Multiplying the inner terms, we have positive two 𝑥. And multiplying the last terms, we have negative 14. On the right-hand side, we just bring down the value of negative eight. Next, we simplify by collecting like terms. Negative seven 𝑥 plus two 𝑥 gives negative five 𝑥. And adding eight to each side, we have a constant term of negative six on the left-hand side. That’s negative 14 plus eight. And we have zero on the right-hand side.

Our quadratic equation is now in its most easily recognizable form. That is the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 for some constants 𝑎, 𝑏, and 𝑐. We’re now going to attempt to solve this quadratic equation by factoring. The coefficient of 𝑥 squared is one, which means that the first term in each of our sets of parentheses must be 𝑥 because 𝑥 multiplied by 𝑥 gives 𝑥 squared. To complete our parentheses, we’re then looking for two numbers whose sum is the coefficient of 𝑥, that’s negative five, and whose product is the constant term, that’s negative six. We can list out the factor pairs of six. They are one and six and two and three.

In order for the product to be negative six though, we need one number to be positive and one to be negative. If we take our factor pair one and six and we make the six negative, then the product is negative six and the sum is indeed negative five. So these are the two numbers we’re looking for to complete our parentheses. So we can fill them in. And we see that the factored form of our quadratic is 𝑥 plus one multiplied by 𝑥 minus six is equal to zero. We can, of course, confirm this by redistributing these parentheses if we wish.

Next, we recall that we have two factors multiplying to give zero. And the only way this can happen is if a least one of the individual factors is itself 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 linear equation. We have 𝑥 plus one is equal to zero or 𝑥 minus six is equal to zero. Subtracting one from each side of the first equation gives 𝑥 equals negative one. And adding six to each side of the second equation gives 𝑥 equals six. Therefore, there are two values in the solution set of this equation, the values negative one and six.

We can, of course, check each of these values by substituting them into the original equation. For example, when 𝑥 equals negative one, substituting this value gives negative one plus two multiplied by negative one minus seven. That’s one multiplied by negative eight, which is indeed equal to negative eight, the right-hand side of the equation, confirming that 𝑥 equals one is a correct value in the solution set of this equation. So we’ve completed the problem. The solution set of the quadratic equation 𝑥 plus two multiplied by 𝑥 minus seven equals negative eight is the set of values negative one, six.

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