SOLUTION: The hypotenuse of right triangle is twice as long as one of the legs and eight inches longer than the other What are the lengths of the sides of the triangle?

Algebra ->  Pythagorean-theorem -> SOLUTION: The hypotenuse of right triangle is twice as long as one of the legs and eight inches longer than the other What are the lengths of the sides of the triangle?      Log On


   

Question 636954: The hypotenuse of right triangle is twice as long as one of the legs and eight inches longer than the other What are the lengths of the sides of the triangle?
Found 2 solutions by reviewermath, graphmatics:
Answer by reviewermath(1029) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+%282x+-+8%29%5E2+=+%282x%29%5E2
x%5E2+%2B+4x%5E2+-+32x+%2B+64+=+4x%5E2
x%5E2+-+32x+%2B+64+=+0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-32x%2B64+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-32%29%5E2-4%2A1%2A64=768.

Discriminant d=768 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--32%2B-sqrt%28+768+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-32%29%2Bsqrt%28+768+%29%29%2F2%5C1+=+29.856406460551
x%5B2%5D+=+%28-%28-32%29-sqrt%28+768+%29%29%2F2%5C1+=+2.14359353944898

Quadratic expression 1x%5E2%2B-32x%2B64 can be factored:
1x%5E2%2B-32x%2B64+=+1%28x-29.856406460551%29%2A%28x-2.14359353944898%29
Again, the answer is: 29.856406460551, 2.14359353944898. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-32%2Ax%2B64+%29

Therefore x = 29.856406460551, disregard the other root because length cannot be negative.
2x = 59.712812921102
2x - 8 = 51.712812921102
The sides of the triangle to 4 decimal places are
29.8564, 51.7128, and 59.7128.

Answer by graphmatics(170) About Me  (Show Source):
You can put this solution on YOUR website!
Let a and b be the sides of the right triangle and c be the hypotenuse. We are given that:
c=2*a
c=b+8
Being a right triangle c%5E2=a%5E2%2Bb%5E2
,putting each of the expressions for c into this equation we get:
%282%2Aa%29%5E2=4%2Aa%5E2=a%5E2%2Bb%5E2
%28b%2B8%29%5E2=b%5E2%2B16%2Ab%2B64=a%5E2%2Bb%5E2
From equation one we get that
a%5E2=b%5E2%2F3 Put this into equation two:
b%5E2%2B16%2Ab%2B64=b%5E2%2F3%2Bb%5E2
b%5E2-48%2Ab-192=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ab%5E2%2Bbb%2Bc=0 (in our case 1b%5E2%2B-48b%2B-192+=+0) has the following solutons:

b%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-48%29%5E2-4%2A1%2A-192=3072.

Discriminant d=3072 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--48%2B-sqrt%28+3072+%29%29%2F2%5Ca.

b%5B1%5D+=+%28-%28-48%29%2Bsqrt%28+3072+%29%29%2F2%5C1+=+51.712812921102
b%5B2%5D+=+%28-%28-48%29-sqrt%28+3072+%29%29%2F2%5C1+=+-3.71281292110204

Quadratic expression 1b%5E2%2B-48b%2B-192 can be factored:
1b%5E2%2B-48b%2B-192+=+1%28b-51.712812921102%29%2A%28b--3.71281292110204%29
Again, the answer is: 51.712812921102, -3.71281292110204. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-48%2Ax%2B-192+%29

b=51.71 So c=59.71 a=59,71/2=29.85