# Unsure of how to solve this right triangle

• Jan 22nd 2013, 11:52 AM
goldbug78
Unsure of how to solve this right triangle
Attachment 26663
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.
• Jan 22nd 2013, 11:55 AM
Plato
Re: Unsure of how to solve this right triangle
Quote:

Originally Posted by goldbug78
Attachment 26663
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 $x^2+(x+3)^2=29$.
• Jan 22nd 2013, 12:03 PM
goldbug78
Re: Unsure of how to solve this right triangle
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.
• Jan 22nd 2013, 12:13 PM
Plato
Re: Unsure of how to solve this right triangle
Quote:

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.

$\\x^2+(x+3)^2=29\\2x^2+6x+9=29\\2x^2+6x-20=0$
• Jan 22nd 2013, 12:27 PM
goldbug78
Re: Unsure of how to solve this right triangle
I got it now, with the answer 2. I kept thinking it was 3x instead of 2x. Thanks for the help!
• Jan 24th 2013, 11:27 PM
ivwiwba35
Re: Unsure of how to solve this right triangle
• Jan 25th 2013, 09:54 AM
pspss
Re: Unsure of how to solve this right triangle
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
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