Question

hi, I have to find the short leg, long leg, and hypotenuse for a right triangle. The short leg = x, the long leg = 1/2x+11 and the hypotenuse is 2x+1. Using a^2+b^2=c^2, I got x^2+(1/2x+11)^2=2x+1^2 but now I'm stumped and I'm not sure what to do next... please help

Answer 1

you should have this

x^2+(1/2x+11)^2=(2x+1)^2

notice the (2x+1)^2

Answer 2

since you're squaring all of 2x+1

Answer 3

ah, sorry, thank you for fixing that! so, x^2+(1/2x+11)^2=(2x+1)^2. my dad and I are trying to solve this but we're not sure how to multiply now

Answer 4

you now need to FOIL out (1/2x+11)^2 and (2x+1)^2

Answer 5

Okay we got 1/4x^2+11x+121 and 4x^2+2x+1

Answer 6

(2x+1)^2 foils out to 4x^2 + 4x + 1

the other one looks good

Answer 7

okay you're right thank you for fixing that

Answer 8

now get everything to one side and simplify/combine like terms

after that, you use the quadratic formula

Answer 9

We're not really sure if we combined it right we got 11/4x^2-7x-120=0

Answer 10

you are correct, now solve that using the quadratic formula

Answer 11

Optional:

Multiply both sides by 4 to clear out the fraction

11/4x^2-7x-120=0

4*(11/4x^2-7x-120)=4*0

11x^2 - 28x - 480 = 0

then use the quadratic formula

Answer 12

okay so plugging that into the quadratic formula we got x=176 and x=-120 and it can't be negative, so the answer for x would be 176?

Answer 13

I'm not getting either of those possible solutions. I'm getting two completely different numbers. So try again

Answer 14

ahh! i forgot to divide it T.T okay, i divided it this time and i got x=8

Answer 15

yep, the final answer is x = 8

Answer 16

oh my gosh ;-; thank you so so so so much!!!

Answer 17

you're welcome