Can some one help with this?: The hypotenuse of a right triangle is 50 inches long. One leg is 4 inch(es) longer than the other. Find the lengths of the legs of the triangle. Round your answers to the nearest tenth of an inch.

I came up with
(x)^2+(x+4)^2=(50)^2
(2x)^2+16=2500 -16
(2x)^2=2484 /2
(x)^2=1242 sqrt
x=35.24 and 39.24
But it saying the answer is incorrect, what am I doing wrong?

Question

Can some one help with this?: The hypotenuse of a right triangle is 50 inches long. One leg is 4 inch(es) longer than the other. Find the lengths of the legs of the triangle. Round your answers to the nearest tenth of an inch. 

I came up with 
(x)^2+(x+4)^2=(50)^2
(2x)^2+16=2500 -16
(2x)^2=2484  /2
(x)^2=1242  sqrt
x=35.24 and 39.24
But it saying the answer is incorrect, what am I doing wrong?

waleed_imtiaz · Answer

@seattle12345  what U have written in the second line.....? when U solve te squares of the first line U will get x2+x2+16+8x=2500..... see ur calculation

anonymous · Answer

See here first line is correct:

$$x^2 + (x+4)^2 = 50^2 $$


Can you tell me what will you get on expanding this:

$$(x + 4)^2 = ??$$

anonymous · Answer

You have mistaken in expanding this:

$$(a + b)^2 = a^2 + b^2 + 2ab$$

Use this and expand that..

anonymous · Answer

@seattle12345 understanding what I am saying or not ??

anonymous · Answer

No problem you will get a quadratic equation:

$$x^2 + x^2 + 16 + 8x = 2500 \implies 2x^2 + 8x - 2500 + 16 = 0$$

$$2x^2 + 8x - 2484 = 0 \implies x^2 + 4x - 1242 = 0$$

anonymous · Answer

You have to go by quadratic formula here:

$$D= b^2 - 4ac = 16 + 4968 \implies D = 4984$$

$$\sqrt{D} = 70.6$$

$$x = \frac{-4 \pm 70.6}{2} \implies x = \frac{-4 + 70.6}{2} \implies x = 33.3$$