SOLUTION: Is this problem possible to solve?
I have tried applying the Pythagorean theorem, quadratic equations etc. but have been unable to come up with a solvable equation.
The diag
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I have tried applying the Pythagorean theorem, quadratic equations etc. but have been unable to come up with a solvable equation.
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Question 1189850: Is this problem possible to solve?
I have tried applying the Pythagorean theorem, quadratic equations etc. but have been unable to come up with a solvable equation.
The diagonal measure of a small TV screen is 1 inch greater than the length and 2 inches
greater than the width. Find the length and width
I would appreciate if anyone can point out something I am missing, or if this is a flawed problem. Found 3 solutions by mananth, ikleyn, greenestamps:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! There ia no flaw . The answer is obvious 4, length 3 width
Let diagonal be x
Length will be (x-1)
Width will be (x-2)
(x-1)^2+(x-2)^2=x^2
Simplify
x^2-2x+1+x^2-4x+4 =x^2
x^2-6x +5 =0
(x-5)(x-1)=0
x=5
x cannot be 1
Let x be the diagonal length, in inches.
Then the legs are (x-1) inches and (x-2) inches.
Pythagorean equation
x^2 = (x-1)^2 + (x-2)^2.
Simplify and find x
x^2 = x^2 - 2x + 1 + x^2 - 4x + 4
0 = x^2 - 6x + 5
Factor
(x-5)*(x-1) = 0
The roots are 5 and 1; but only 5 gives the solution to the problem.
So, the hypotenuse is 5 inches; the legs are 5-1 = 4 inches and 5-2 = 3 inches.
The dimensions of the small TV screen are 3 inches by 4 inches.
