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madsmckell
word problem!! A parcel of land is 6ft longer than it is wide. Each diagonal from one corner to the opposite corner is 174 ft long. What are the dimensions of the parcel?
Draw a picture. That will help you figure out what to do.
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thanks but how do I solve it?
It is 6 feet longer than it is wide, so you can write an expression for both the length and the width.
okay.... i did that... then what?:)
Use the Pythagorean theorem.
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so... (w+6)^2+w^2=174^2??????
You can do it! I have faith in you!!
i was trying to solve for w but i got stuck... will someone help me and walk me through it??
Stuck where? Show your steps and I will help you?
(w+6)^2+w^2=30276..... w^2+12w+36+w^2=30276......2w^2+12w+36=30267
i was trying to use the quadratic formula but i wasnt sure what to do and my numbers were wayyyyyy big
Great job! Find a calculator, you are going to need it. :-) The quadratic formula is what you will need. I'm still working through it. Before you get to that, get all terms to the left side of the equation and set it equal to 0.
The numbers do get big, but you can handle it, with your calculator's help.
when i do the quadratic inside the root is negative so how would i simplify it??? isnt the quadratic formula -b plus or minus the root of b^2-4ac all over 2a
It is not negative. You dropped a negative on the floor....look around, I'm sure it is still nearby. :-)
One you get all the terms to the left side, the constant term will be negative.
I divided the whole equation by 2 to keep the numbers smaller. \(w^2+6w-15120\)
but when you put in -15120 it is negative in the root soo it wont workkk
the answer in the back of the book is 120 by 126
so still not sure how that works when what ever i root comes out to be a decimal
The answer in the back of the book is correct. \[\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\] \[w^2+6w-15120\] \[\dfrac{-6 \pm \sqrt{6^2-4(1)(-15120}}{2(1)}\] remember that a negative times a negative is a positive.
wow thanks soooo mcuh!! you are AMAZZZZINNNGGG!!!!