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.

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- Jan 22nd 2013, 10:52 AMgoldbug78Unsure 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, 10:55 AMPlatoRe: Unsure of how to solve this right triangle
- Jan 22nd 2013, 11:03 AMgoldbug78Re: 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, 11:13 AMPlatoRe: Unsure of how to solve this right triangle
- Jan 22nd 2013, 11:27 AMgoldbug78Re: 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, 10:27 PMivwiwba35Re: Unsure of how to solve this right triangle
thank you for your article,My problem has been resolved.

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Cheap North Face Jackets - Jan 25th 2013, 08:54 AMpspssRe: 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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