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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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
Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.

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.

Solved.



Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


Let the diagonal length be x
Then the length is x-1 and the width is x-2

Apply the Pythagorean Theorem:






or

x=1 doesn't work; it makes the length 0 and the width -1, which is nonsense.

x=5 works. The length is x-1=4 and the width is x-2=3; then 3^2+4^2=5^2
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