# Passage of a norm of a vector exercise

• Dec 1st 2008, 07:53 PM
pedrosacosta
Passage of a norm of a vector exercise
Hi,

I've an example of vector exercise that achieved the following step onwards to resolving. |vector r| is the norm of the vector.

|vector r|^2 = 100(25t^2-14t+2) = 100((5t-(14/10)^2) - (14/10)^2 +2)

I don't understand how the passage is made, because for me the values inside the parameter of the first count is a quadratic function who has the consequent result:
t=(-14+/-squarert(-4))/50.

Can anybody explain me how the passage of the |vector r|^2 is made?

Thanks,
Pedro
• Dec 2nd 2008, 10:34 AM
running-gag
Hi Pedro

$\displaystyle (5t-\frac{14}{10})^2\;=\;25t^2-14t+(\frac{14}{10})^2$

Therefore
$\displaystyle 25t^2-14t+2\;=\;(5t-\frac{14}{10})^2-(\frac{14}{10})^2+2$

• Dec 2nd 2008, 11:24 AM
earboth
Quote:

Originally Posted by pedrosacosta
Hi,

I've an example of vector exercise that achieved the following step onwards to resolving. |vector r| is the norm of the vector.

|vector r|^2 = 100(25t^2-14t+2) = 100((5t-(14/10))^2 - (14/10)^2 +2)

I don't understand how the passage is made, because for me the values inside the parameter of the first count is a quadratic function who has the consequent result:
t=(-14+/-squarert(-4))/50.

Can anybody explain me how the passage of the |vector r|^2 is made?

Thanks,
Pedro

You've made an insidious typo: See the correction in red.

Here is nothing done but completing the square:

$\displaystyle |\vec r|^2=100(25t^2-14t+2)=100((5t)^2-14t+\left(\frac{14}{10}\right)^2 - \left(\frac{14}{10}\right)^2+2) =$ $\displaystyle 100\left(\left(5t-\left(\frac{14}{10}\right)\right)^2-\left(\frac{14}{10}\right)^2+2\right)$

You only have to translate the ^2 behind the closing paranthese.