Yes, use the quadratic formula, you'll find two different zero's and therefore the quadratic equation can be factorized as (in general):
Can you do that?
If then there are two real solutions but it doesn't state whether or not they're rational.
If it's a square number then it will factor over the rational numbers which is generally why equations are factored.
That means it won't factor over the rational numbers. The discriminant is positive so there are still two real solutions.Originally Posted by OP
I am actually working with a long word problem, so I will give you guys the word problem. I am positive there should have been a minus sign. I know the correct answer to this problem. I just can't make sense of it.
Info: A park is going to be made on unused land that a city owns. The park will be a rectangular region 60 feet by 150 feet with an area of 9,000 square feet.
Problem: The long term plan for the park involves doubling the area of the park. The length and width of the park will each be extended by d feet. For which of the following equations is x=d a solution?
A. (x + 60)(x + 150) = 2(9,000) This is the correct answer.
I have mused over this problem for over 20 minutes (obviously to no avail). I think I am fundamentally misunderstanding the question.
I still don't understand how/why it is the correct answer.
I'm confused about the x=d solution. Isn't x the same as D for both the length and width?
Wouldn't x have to be a rational number to qualify as a solution?
It appears that you had a number of equations to choose from. This one DOES describe the solution to the problem. (If you solve for x and then let d = x, where x is that solution, then adding that value of d to both the length & width will double the area of the park.
There is nothing in this problem that says d has to be rational. In fact, no rational value of d will work. (any rational value for d that's greater than 15 sqrt(89)-105 will more than double the area.)
BTW: It also appears that the intent of the exercise was to see if you could set-up an equation which could be used to solve the problem, rather than to actually follow through with the solution.