Question 179318
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 18 inches more than the diagonal measure of the smaller. What is the height of the screens, correct to the nearest 0.1 inch?
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Let x = height of both screens:
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Conventional screen dimensions: (x+3 by x
HD screen dimensions: 1.8x by x
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Find x by writing an equation of the diagonals (hypotenuse) of the screens
HD diagonal = Conventional diagonal + 18 inches
{{{sqrt(x^2 + (1.8x)^2)}}}  = {{{sqrt(x^2 + (x+3)^2)}}} + 18
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{{{sqrt(x^2 + 3.24x^2)}}}  = {{{sqrt(x^2 + (x^2+6x+9))}}} + 18
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{{{sqrt(4.24x^2)}}} - {{{sqrt(2x^2+6x+9))}}} - 18 = 0
To avoid a tremendous amt of math, graph this equation:
{{{ graph( 300, 200, -2, 50, -20, 20, sqrt(4.24x^2) - sqrt(2x^2+6x+9)- 18) }}}
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Note the positive intercept occurs at x~31" or 31.3" on a Ti83
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Check solution by finding the diagonals using x = 31.3 
HD {{{sqrt(31.3^2 + (1.8*31.3)^2)}}} = 64.4
Convent {{{sqrt(31.3^2 + 34.3^2)}}} =  46.4
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This will give us a difference of:    18.0" 
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