1. ## [SOLVED] pythagoras

hi,

having trouble seeing where i'm going wrong.

c^2\=\a^2\+\b^2

c\=\13,\a\=\3x,\b\=\3x

13^2\=\(3x)^2\+\(2x)^2

169\=\6x^2\+\4x^2

169\=\10x^2

16.9\=\x^2\

x\=\pm\sqrt{16.9}

x\=\sqrt{16.9}

text book hast the answer as sprt{13}

How?

btw, I wrap math tags around these selected text and latex/syntex error comes up.

anyone?

2. Originally Posted by nativecat
hi,

having trouble seeing where i'm going wrong.

c^2\=\a^2\+\b^2

c\=\13,\a\=\3x,\b\=\3x

13^2\=\(3x)^2\+\(2x)^2

169\=\6x^2\+\4x^2 ... (3x)^2 = 9x^2, not 6x^2

169\=\10x^2

16.9\=\x^2\

x\=\pm\sqrt{16.9}

x\=\sqrt{16.9}

text book hast the answer as sprt{13}

How?

btw, I wrap math tags around these selected text and latex/syntex error comes up. because the back slash {\} is only necessary for certain things.

anyone?
.

3. Originally Posted by nativecat
hi,

having trouble seeing where i'm going wrong.

c^2\=\a^2\+\b^2

c\=\13,\a\=\3x,\b\=\3x

13^2\=\(3x)^2\+\(2x)^2

169\=\6x^2\+\4x^2

169\=\10x^2

16.9\=\x^2\

x\=\pm\sqrt{16.9}

x\=\sqrt{16.9}

text book hast the answer as sprt{13}

How?

btw, I wrap math tags around these selected text and latex/syntex error comes up.

anyone?
The answers are $x = \pm \sqrt{13}$ because you can have either a positive square root or a negative one. If you're talking about a length it will always be positive when dealing with scalars

Skeeter pointed out that $(3x)^2 = 3^2 \times x^2 = 9x^2$