Question 249403
The diagonal of the TV can be considered the hypotenuse of a right triangle.  It can be solved using the Pythagorean formula:  
{{{C^2 = A^2 + B^2}}} 
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Assuming L is the width of the TV set, which the problem calls "length", and H is the height of the screen.
{{{L = H + 28}}}
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Returning to the Pythagorean formula...
{{{D = diagonal = 52}}}
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{{{D^2 = L^2 + H^2}}}
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Substituting L = H+28
{{{D^2 = (H+28)^2 + H^2}}}
{{{52^2 = (H+28)(H+28) + H^2}}}
{{{52^2 = H^2 + 28H + 28H + 784 + H^2}}}
{{{52^2 = 2H^2 + 56H + 784}}}
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Squaring 52
{{{2704 = 2H^2 + 56H + 784}}}
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Dividing both sides by 2
{{{1352 = H^2 + 28H + 392}}}
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Subtracting 1352 from both sides
{{{0 = H^2 + 28H + 392 - 1352}}}
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Simplifying
{{{0 = H^2 + 28H -960}}}
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Can 960 be factored such that the two terms are 28 apart?
Yes.  48*20 = 960 and 48-20=28
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{{{(H+48)(H-20) = 0}}}
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So we have two candidate solutions:  H= -48 and H = 20.  Since a negative height is nonsense, then our suggested answer is:
{{{H = 20}}}
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Looking back to our defined equations, 
{{{L = H + 28 = 20 + 28 = 48}}}
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Checking by using the Pythagorean formula:
{{{D = 52}}} 
{{{L^2 = 48^2 = 2304}}}
{{{H^2 = 20^2 = 400}}}
{{{L^2 + H^2 = 2704}}}
{{{sqrt(2704) = 52}}}
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So that checks just fine.
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But is it an analog TV or an HDTV?
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Analog TV has a ratio of width to height of 4:3.  The picture is 4 units wide by 3 units high.
HDTV has a ration of 16:9.  The picture is 16 units wide by 9 units high.
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Our proposed TV set has a picture that is 48 wide by 20 high.  That is a ratio of 48:20, or 24:10, or 12:5.  This does not correspond to any real TV set.  So perhaps a negative height would work when solving an "imaginary" TV problem. Hmmm...