# Can I get help with this triq problem?

• Jun 9th 2008, 04:14 PM
Darkhrse99
Can I get help with this triq problem?
I'm trying to find X P( x, -2/7)

$\displaystyle x^2+ -2/7=1$ then I get
$\displaystyle x^2+ (-2/7)^2=1$
Than I get $\displaystyle x^2=1-4/49=$ $\displaystyle x^2= \frac{49} {49}- \frac{4}{49}=\frac{45} {49}$ Then you get $\displaystyle x=\sqrt \frac{45} {49}$The answer is suppose to be $\displaystyle x=+ or -\sqrt{5}$.

How do you get $\displaystyle x= + - \sqrt{5}$

Thanks

Jason
• Jun 9th 2008, 04:54 PM
topsquark
Quote:

Originally Posted by Darkhrse99
I'm trying to find X P( x, -2/7)

$\displaystyle x^2+ -2/7=1$ then I get
$\displaystyle x^2+ (-2/7)^2=1$
Than I get [tex]x^2=1-4/49= $\displaystyle x^2= 49/49-4/49=x^2=45/49$ The answer is suppose to be[tex] x=\sqrt{5}.

How do you get sqrt + - 5?

Thanks

Jason

Can I just be the first to say

Huh? (Worried)

You started with
$\displaystyle x^2 \pm \frac{2}{7} = 1$

and went to
$\displaystyle x^2 + \left ( -\frac{2}{7} \right )^2 = 1$

How did you go from the first to the second step? The first does not imply the second.

What exactly is it you are trying to find? "X P (x, -2/7)" makes no sense to me.

-Dan
• Jun 9th 2008, 05:36 PM
Darkhrse99
Sorry for the confusion, the math code is kicking my butt.

I'm trying to find point X on the unit circle.

The book gave me the Y cord of $\displaystyle \frac {-2} {7}$ . I need to find X.

The formula $\displaystyle x^2+ y^2 =1$ is used to find the answer.

I plugged in $\displaystyle x^2 + \frac {-2} {7}squared=1$

Than I get $\displaystyle x^2=1-\frac{4} {49} next I get$ $\displaystyle x^2= \frac{49} {49}- \frac{4}{49}=\frac{45} {49}$ Then you get $\displaystyle x=\sqrt \frac{45} {49}$The answer is suppose to be $\displaystyle x=+ or -\sqrt{5}$.
• Jun 9th 2008, 05:57 PM
Darkhrse99
I found the answer. The book was wrong. The answer is $\displaystyle 3\sqrt{5}/7$
• Jun 9th 2008, 06:20 PM
topsquark
Quote:

Originally Posted by Darkhrse99
Sorry for the confusion, the math code is kicking my butt.

I'm trying to find point X on the unit circle.

The book gave me the Y cord of $\displaystyle \frac {-2} {7}$ . I need to find X.

The formula $\displaystyle x^2+ y^2 =1$ is used to find the answer.

I plugged in $\displaystyle x^2 + \frac {-2} {7}squared=1$

Than I get $\displaystyle x^2=1-\frac{4} {49} next I get$ $\displaystyle x^2= \frac{49} {49}- \frac{4}{49}=\frac{45} {49}$ Then you get $\displaystyle x=\sqrt \frac{45} {49}$The answer is suppose to be $\displaystyle x=+ or -\sqrt{5}$.

The solution is
$\displaystyle \pm \frac{3 \sqrt{5}}{7}$
Yes, mostly right. I was stumped all day. The solutions manual said $\displaystyle \pm\sqrt{5}$ but the text book said $\displaystyle \pm \frac{3 \sqrt{5}}{7}$