SOLUTION: Two television monitors sitting beside each other on a shelf in an appliance store have the same screen height. One has a conventional screen, which is 3 inches wider than it is hi

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Question 157758: Two television monitors sitting beside each other on a shelf in an appliance store have the same screen height. One has a conventional screen, which is 3 inches wider than it is high. The other has a wider, high-definition screen, which is 1.8 times as wide as it is high. The diagonal measure of the wider screen is 15 inches more than the diagonal measure of the smaller. What is the height of the screens, correct to the nearest 0.1 inch?
I tried the Pythagorean theorem & got 2w^2+6w+24=d^2 and got stuck.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You have to keep track of both sets of heights and widths then compare diagonals.
For the conventional TV,
w%5Bc%5D=h%5Bc%5D%2B3
For the high-def TV,
w%5Bh%5D=1.8h%5Bh%5D
The diagonals of each would be,
Conventional
d%5Bc%5D%5E2=w%5Bc%5D%5E2%2Bh%5Bc%5D%5E2
d%5Bc%5D%5E2=%28h%5Bc%5D%2B3%29%5E2%2Bh%5Bc%5D%5E2
d%5Bc%5D%5E2=%28h%5Bc%5D%5E2%2B6h%5Bc%5D%2B9%29%2Bh%5Bc%5D%5E2
d%5Bc%5D%5E2=%28h%5Bc%5D%5E2%2B6h%5Bc%5D%2B9%29%2Bh%5Bc%5D%5E2
d%5Bc%5D%5E2=2h%5Bc%5D%5E2%2B6h%5Bc%5D%2B9
d%5Bc%5D=sqrt%282h%5Bc%5D%5E2%2B6h%5Bc%5D%2B9%29
High def
d%5Bh%5D%5E2=w%5Bh%5D%5E2%2Bh%5Bh%5D%5E2
d%5Bh%5D%5E2=%281.8h%5Bh%5D%29%5E2%2Bh%5Bh%5D%5E2
d%5Bh%5D%5E2=3.24h%5Bh%5D%5E2%2Bh%5Bh%5D%5E2
d%5Bh%5D%5E2=4.24h%5Bh%5D%5E2
d%5Bh%5D=2.059h%5Bh%5D
You also know that
d%5Bh%5D=d%5Bc%5D%2B15
and that
h%5Bh%5D=h%5Bc%5D=h
We can then substitute,
d%5Bh%5D=d%5Bc%5D%2B15
%282.059h%29=%28sqrt%282h%5E2%2B6h%2B9%29%29%2B15
2.059h-15=sqrt%282h%5E2%2B6h%2B9%29
%282.059h-15%29%5E2=2h%5E2%2B6h%2B9
4.239h%5E2-61.77h%2B225=2h%5E2%2B6h%2B9
2.239h%5E2-67.77h%2B216=0
Use the quadratic formula,
h+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

h+=+%2867.77+%2B-+sqrt%28+4592.77-1934.50%29%29%2F%284.478%29
h+=+%2867.77+%2B-+sqrt%28+2658.27%29%29%2F%284.478%29
h+=+%2867.77+%2B-+51.558%29%2F%284.478%29
h+=+%2867.77+%2B+51.558%29%2F%284.478%29
h+=+%28119.328%29%2F%284.478%29
h=26.6
Check the answer
h%5Bc%5D=26.6
w%5Bc%5D=29.6
d%5Bc%5D=sqrt%2826.6%5E2%2B29.6%5E2%29
d%5Bc%5D=39.796
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h%5Bh%5D=26.6
w%5Bh%5D=47.88
d%5Bh%5D=sqrt%2826.6%5E2%2B47.88%5E2%29
d%5Bh%5D=54.772
d%5Bh%5D-d%5Bc%5D=54.772-39.796=14.976
Close enough to 15.
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A curiosity to consider (or ask the teacher) is the other solution from the quadratic formula.
h+=+%2867.77+%2B-+51.558%29%2F%284.478%29
h+=+%2867.77+-+51.558%29%2F%284.478%29
h+=+%2816.212%29%2F%284.478%29
h+=+3.62
From this height then,
h%5Bc%5D=3.62
w%5Bc%5D=6.62
d%5Bc%5D=sqrt%283.62%5E2%2B6.62%5E2%29
d%5Bc%5D=56.93
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h%5Bh%5D=3.62
w%5Bh%5D=6.52
d%5Bh%5D=sqrt%283.62%5E2%2B6.52%5E2%29
d%5Bh%5D=55.62
d%5Bh%5D-d%5Bc%5D=56.93-55.62=1.31
This height does not satisfy the diagonal requirement.
Why? Good question to ask the teacher.