SOLUTION: The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures √6m. Find the width and height.

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Question 78137: The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures √6m. Find the width and height.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let h=height, w=width
So we have
w=2%2Bh

Since the diagonal is the hypotenuse, we can say
w%5E2%2Bh%5E2=%28sqrt%286%29%29%5E2

%282%2Bh%29%5E2%2Bh%5E2=%28sqrt%286%29%29%5E2 Substitute 2+h into w

4%2B4h%2Bh%5E2%2Bh%5E2=%28sqrt%286%29%29%5E2 foil the parenthesis

4%2B4h%2B2h%5E2=6 Combine like terms and reduce %28sqrt%286%29%29%5E2 to 6

4%2B4h%2B2h%5E2-6=0 Subtract 6 from both sides

2h%5E2%2B4h-2=0 Combine like terms
Now use the quadratic formula to find the value of L:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ah%5E2%2Bbh%2Bc=0 (in our case 2h%5E2%2B4h%2B-2+=+0) has the following solutons:

h%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=%284%29%5E2-4%2A2%2A-2=32.

Discriminant d=32 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+32+%29%29%2F2%5Ca.

h%5B1%5D+=+%28-%284%29%2Bsqrt%28+32+%29%29%2F2%5C2+=+0.414213562373095
h%5B2%5D+=+%28-%284%29-sqrt%28+32+%29%29%2F2%5C2+=+-2.41421356237309

Quadratic expression 2h%5E2%2B4h%2B-2 can be factored:
2h%5E2%2B4h%2B-2+=+2%28h-0.414213562373095%29%2A%28h--2.41421356237309%29
Again, the answer is: 0.414213562373095, -2.41421356237309. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B4%2Ax%2B-2+%29


Ignoring the negative answer we get:

h=0.414
So use this to find w
w=2%2B0.414=2.414
So the height is 0.414 m and the width is 2.414 m approximately