How do I calculate the following 1- √(〖(0-PF)〗^2+ 〖(1-PD)〗^2 )/√2

Guest Aug 8, 2017
social bar

2+0 Answers



How do I calculate the following 1- √(〖(0-PF)〗^2+ 〖(1-PD)〗^2 )/√2


\(1-\frac {\sqrt{(0-PF)^2+(1-PD)^2}}{\sqrt{2}}\)

\(=1-\frac {\sqrt{(PF)^2+(1-2PD+(PD)^2)}}{\sqrt{2}}\)

\(=1-\frac {\sqrt{P^2F^2+1-2PD+P^2D^2}}{\sqrt{2}}\)

\(=1-\frac {\sqrt{P^2F^2+1-2PD+P^2D^2}}{\sqrt{2}}\)


\(=1-\frac {\sqrt{2P^2F^2+2-4PD+2P^2D^2}}{2}\)


laugh  !  Thanks \(\large{X^2}\)

asinus  Aug 8, 2017
edited by asinus  Aug 8, 2017

Let's see if I can simplify this expression of \(1-\frac{\sqrt{(0-PF)^2+(1-PD)^2}}{\sqrt{2}}\)

\(1-\frac{\sqrt{(0-PF)^2+(1-PD)^2}}{\sqrt{2}}\) Let's first notice that 0 minus a number is always itself.
\(1-\frac{\sqrt{(PF)^2+(1-PD)^2}}{\sqrt{2}}\) Let's do \((PF)^2\) by using the exponent rule that \((ab)^n=a^n*b^n\)
\(1-\frac{\sqrt{P^2F^2+(1-PD)^2}}{\sqrt{2}}\) Now, let's expand (1-PD)^2 by using the exponent rule that \((a-b)^2=a^2-2ab+b^2\)
\(1^2-2(1)(PD)+(PD)^2\) Simplify 
\(1-2PD+P^2D^2\) Now, reinsert this into the equation for \((1-PD)^2\).
\(1-\frac{\sqrt{P^2F^2+1-2PD+P^2D^2}}{\sqrt{2}}\) Rearrange the terms in the numerator such that the terms are ordered, from left to right, based on degree.
\(1-\frac{\sqrt{P^2F^2+P^2D^2-2PD+1}}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}\) Now, I want to deal with the denominator. Rationalize it by multiplying by \(\frac{\sqrt{2}}{\sqrt{2}}\). To combine the numerator and denominator, note that \(\sqrt{a}*\sqrt{b}=\sqrt{ab}\)
\(1-\frac{\sqrt{2(P^2F^2+P^2D^2-2PD+1)}}{2}\) There is no common factor amongst the terms in the numerator other then 2, so there is no more simplification that can be done. 
TheXSquaredFactor  Aug 8, 2017

14 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details