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

Guest Aug 8, 2017


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

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