Question 942348: 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 in. wider than it is high. The other has a wider, high-definition screen, which is 1.9 times as wide as it is high. The diagonal measure of the wider screen is 16 in. more than the diagonal measure of the smaller. What is the height of the screens, correct to the nearest 0.1 in.?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the height is equal to 24.8 rounded to the nearest tenth.
you have 2 equations:
from the standard size screen, you get x^2 + (x+3)^2 = y^2
from the wide screen, you get x^2 + (1.9x)^2 = (y+16)^2
from the second equation, you solve for y^2 to get y^2 = (sqrt(4.61x^2) - 16)^2
you can then substitute for y^2 in the first equation to get:
x^2 + (x+3)^2 = (sqrt(4.61x^2) - 16)^2
you then simplify to get:
x^2 + x^2 + 6x + 9 = 4.61x^2 - 32*sqrt(4.61x^2) + 256
you then simplify further to get:
2x^2 + 6x + 9 = 4.61x^2 - 32*sqrt(4.61)*x + 256
you then simplify further to get:
2x^2 + 6x + 9 = 4.61x^2 - 68.70691377x + 256
you then subtract (2x^2 + 6x + 9) from both sides of the equation to get:
0 = 4.61x^2 - 2x^2 - 68.70691377x - 6x + 256 - 9
you then combine like terms to get:
0 = 2.61x^2 - 74.70691377x + 247.
this is a quadratic equation that is in standard form.
you get:
a = 2.61
b = -74.70691377
c = 247
you use the quadratic formula to get:
x = 24.80870998 or x = 3.814628628.
you then place both values in the original equations to see if they're true.
x = 3.814628628 doesn't work so it's extraneous and is discarded.
x = 24.80870998 works so that's your solution.
when x = 24.80870998:
y = square root of (x^2 + (x+3)^2) = 37.2665593 and (y+16) = square root of (x^2 + (1.9x)^2) = 53.2665593.
the difference between them is 16 per the requirements of the problem.
round to the nearest tenth and 24.8 is your solution.
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