Thread: Unsure of how to solve this right triangle

Thats 29 squared on the hypotenuse and x+3 on the opposite and just x on the adjacent. I cant figure what to do with the 29 after it is given an exponent of 2 and taken out of square form.

Originally Posted by goldbug78

Thats 29 squared on the hypotenuse and x+3 on the opposite and just x on the adjacent. I cant figure what to do with the 29 after it is given an exponent of 2 and taken out of square form.

You need to solve .

yeah, but what do with 3x+6x+9=29?I am not sure how to solve this quadratic expression. I know 3x is raised to the second power, just can't figure how to use the tools on this site yet.

Originally Posted by goldbug78

yeah, but what do with 3x+6x+9=29?I am not sure how to solve this quadratic expression. I know 3x is squared, just can't figure how to use the tools on this site yet.

I got it now, with the answer 2. I kept thinking it was 3x instead of 2x. Thanks for the help!

thank you for your article,My problem has been resolved.
Sacs Longchamp
Sacs Longchamp Hobo
Doudoune Armani Homme
Doudoune North Face Femme
Doudoune Calvin Klein Homme
Cheap North Face Jackets

x² + (x + 3)² = (√29)²
x² + x² + 6x + 9 = 29
2x² + 6x – 20 = 0
x² + 3x – 10 = 0
x = [– 3 ± √(3² + 40)]/2
= [– 3 ± √49]/2
= (– 3 ± 7)/2
= taking + ve values
x = (– 3 + 7)/2 = 2
sides are 2, 5, √29 units
----