SOLUTION: please help me I can up with three crazy answers. Translate the following into a quadratic equation, and solve it: The length of a rectangular garden is three times its width;

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: please help me I can up with three crazy answers. Translate the following into a quadratic equation, and solve it: The length of a rectangular garden is three times its width;       Log On


   

Question 172398This question is from textbook
: please help me I can up with three crazy answers.
Translate the following into a quadratic equation, and solve it: The length of a rectangular garden is three times its width; if the area of the garden is 75 square meters, what are its dimensions?
also can you show me how to check the answer after solving the problem.
 This question is from textbook

Found 2 solutions by jim_thompson5910, Mathtut:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since "The length of a rectangular garden is three times its width", this means that L=3w

Area: A=L%2AW


A=L%2Aw Start with the area formula


75=%283w%29%2Aw Plug in A=75 (the given area) and L=3w


75=3w%5E2 Multiply


3w%5E2=75 Rearrange the equation.


w%5E2=%2875%29%2F%283%29 Divide both sides by 3 to isolate z.


w%5E2=25 Reduce.


w=5 Take the square root of both sides (only the positive square root is considered since a negative width doesn't make sense)


So the width is 5 m and the length (which is 3 times the width) is 3*5=15 m


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Answer:


So the width is 5 m and the length is 15 m

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
ok so the we have Width W and Length L=3W....we also know that area=LW
:
3W(W)=75
:
3W%5E2=75
:
3W%5E2-75=0this would be easy enough to solve without the quadratic formula but since you asked for it
:
system%28W=5%2CW=-5%29
since we know we cant have negative lengths
highlight%28W=5%29 so
highlight%28L=3%285%29%29=15
:
checking 3%285%29%5E2=75=3%2825%29=75..and 75=75
:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aW%5E2%2BbW%2Bc=0 (in our case 3W%5E2%2B0W%2B-75+=+0) has the following solutons:

W%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%280%29%5E2-4%2A3%2A-75=900.

Discriminant d=900 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+900+%29%29%2F2%5Ca.

W%5B1%5D+=+%28-%280%29%2Bsqrt%28+900+%29%29%2F2%5C3+=+5
W%5B2%5D+=+%28-%280%29-sqrt%28+900+%29%29%2F2%5C3+=+-5

Quadratic expression 3W%5E2%2B0W%2B-75 can be factored:
3W%5E2%2B0W%2B-75+=+3%28W-5%29%2A%28W--5%29
Again, the answer is: 5, -5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B0%2Ax%2B-75+%29



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