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September 20, 2014

September 20, 2014

Posted by **purpletree** on Sunday, March 29, 2009 at 12:58pm.

- Quadratic Equations -
**Brianna**, Sunday, March 29, 2009 at 2:58pmshoter leg = x

longer leg = 3x+3

hypot= 4x-3

to solve for hypotenuse you use

a^2 + b^2 = c^2, so

(x)^2 + (3x+3)^2 = (4x-3)^2

(x)(x) + (3x+3)(3x+3) = (4x-3)(4x-3)

then multiply

(x^2) + (9x^2+18x+9) = (16x^2-24x+9)

then combine like terms

10x^2 + 18x + 9 = 16x^2 - 24x + 9

then i subtracted 10x^2 from both sides

and got:

18x + 9 = 6x^2 - 24x +9

next, i subtracted 18x from both sides:

9 = 6x^2 - 42x + 9

and then subtracted 9 from both sides

0 = 6x^2 - 42x

then pull out a 6x on the right side of the equation

0 = 6x(x-7)

so

6x=0

and

x-7=0

so x= 0 and 7

so the shorter leg is :

0 units or 7 units

the longer leg is :

3 units or 24 units

and the hypotenuse is :

-3 units or 25

and since you cant have a hypotenuse with a negative value

i think the three lengths of your triangle are

7, 24, and 25

- Quadratic Equations -
**purpletree**, Monday, March 30, 2009 at 9:22amThanks for the help. :)

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