Question

The three sides of a right triangle are x, x+3, and 2x-3.( 11 years ago

Answer 1

ok, now we know what the three sides are
I wouldn't be losing sleep over that now

Answer 2

well i need to find x

Answer 3

read the whole question @jdoe0001 ^^^^

Answer 4

hmm that was added just now

Answer 5

well anyways, what is x?

Answer 6

#phythagorean theorem

Answer 7

\(\bf \textit{pythagorean}=c^2=a^2+b^2\implies c=\pm \sqrt{a^2+b^2}
\\ \quad \\
2x-1=\pm \sqrt{(x)^2+(x+3)^2}\)

solve for "x"

Answer 8

how do i solve for x?

Answer 9

expand the binomial on the left-side, and the one on the right-side
then simplify

Answer 10

What to you mean by expand? Do the multiplication so it's 4x+1?

Answer 11

pythagorean theorem I meant anyhow
you'd use the pythagorean theorem, since it's a right-triangle

Answer 12

hmmm

Answer 13

shoot I got those vlaues a bit off... .lemme redo that

Answer 14

would it be 4x+1?

Answer 15

\(\bf \large {
\textit{pythagorean theorem}=c^2=a^2+b^2
\\ \quad \\
(2x-3)^2=(x)^2+(x+3)^2 }\)

Answer 16

4x+1=x^2+2x+9?

Answer 17

hmmm actually you'd need to use FOIL or just multply the binomials
bear in mind that \(\bf (2x-3)^2=(x)^2+(x+3)^2\implies (2x-3)(2x-3)=(x)^2+(x+3)(x+3)\)

Answer 18

hmm got a bit truncated.... ok so \(\bf (2x-3)^2=(x)^2+(x+3)^2
\\ \quad \\
\implies (2x-3)(2x-3)=(x)^2+(x+3)(x+3)
\)

Answer 19

ok so i got 4x^2-12x+9=x^4+6x+9

Answer 20

is that correct?

Answer 21

@jdoe0001

Answer 22

hmm close one sec

Answer 23

hmmm wait a sec.. shoot... lemme fix that quick

Answer 24

\(\bf (2x-3)^2=(x)^2+(x+3)^2
\\ \quad \\
(2x-3)(2x-3)=(x)^2+(x+3)(x+3)
\\ \quad \\
4x^2-12x+9=x^2+x^2+6x+9
\\ \quad \\
4x^2-12x+9=2x^2+6x+9
\\ \quad \\
2x^2-18x=0\implies 2x(x-9)=0\impliedby common\ factoring
\\ \quad \\
2x(x-9)=0\implies
\begin{cases}
2x=0\to x=\frac{0}{2}\to &x=0\\
x-9=0\to &x=9
\end{cases}\)

Answer 25

so x is 9, so the sides are 9, 12, and 15?

Answer 26

yeap

Answer 27

Ok thanks

Answer 28

yw