Question Video: Solving Word Problems by Factoring a Quadratic Equation | Nagwa Question Video: Solving Word Problems by Factoring a Quadratic Equation | Nagwa

Question Video: Solving Word Problems by Factoring a Quadratic Equation Mathematics • Second Year of Preparatory School

Given two positive numbers whose product is 35 and whose difference is 2, find the largest number.

04:29

Video Transcript

Given two positive numbers whose product is 35 and whose difference is two, find the largest number.

Let’s begin by introducing the letter 𝑛 to represent the larger number. We could use a different letter to represent the other number. But instead, let’s see if we can write an expression for the smaller number in terms of 𝑛. We’re told that the difference between the two numbers is two. It therefore follows that if 𝑛 represents the larger number, the smaller number is equal to 𝑛 minus two. As both numbers must be positive, we have that 𝑛 and 𝑛 minus two must be greater than zero, which leads to 𝑛 is greater than two.

The other information we’re given is that the product of the two numbers is 35. We can use this to form an equation involving the expressions we wrote for the two numbers. 𝑛 multiplied by 𝑛 minus two is equal to 35. Distributing the parentheses on the left-hand side gives 𝑛 squared minus two 𝑛 is equal to 35. And then, we can group all the terms on the same side of the equation by subtracting 35 from each side, giving 𝑛 squared minus two 𝑛 minus 35 equals zero.

We now observe that we have a quadratic equation in 𝑛. We may be able to solve this equation by factoring. If this equation can be factored, then its factored form will be 𝑛 plus 𝑎 multiplied by 𝑛 plus 𝑏 equals zero, for two values 𝑎 and 𝑏 which we need to determine. The numbers 𝑎 and 𝑏 must satisfy two properties. Their sum must be equal to the coefficient of 𝑛 in the quadratic equation, which is negative two. Their product must be equal to the constant term, which is negative 35. As the product is negative, we can deduce that 𝑎 and 𝑏 must have different signs. The pair of numbers that satisfy these properties is negative seven and five. It doesn’t matter which we choose to be 𝑎 and which we choose to be 𝑏.

We therefore have that the factored form of the quadratic equation is 𝑛 minus seven multiplied by 𝑛 plus five is equal to zero. We can check this factored form is correct by redistributing the parentheses if we wish.

Now we need to use this factored form to solve the equation. To do this, we need to recall that if the product of two values is equal to zero, at least one of those values must itself be equal to zero. We therefore set each of the linear factors equal to zero, giving the two linear equations 𝑛 minus seven equals zero or 𝑛 plus five equals zero. We can solve each of these equations in one step. Adding seven to each side of the first equation gives 𝑛 equals seven. And subtracting five from each side of the second equation gives 𝑛 equals negative five.

Now, whilst these are both valid solutions to the quadratic equation, they aren’t both valid in the context of the problem. Recall that 𝑛 represents the larger of the two numbers. And as both 𝑛 and the number two less than 𝑛 must be positive, we stated that 𝑛 must be greater than two. As a result, we can disregard the solution 𝑛 equals negative five because it doesn’t satisfy this condition.

We’ve therefore found that the larger number is seven. It then follows that the smaller number is five. The difference between these two numbers is two, and their product is 35. So this confirms our answer is correct. We’ve found that the larger of the two numbers is seven.

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