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# Expression

0
385
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How do I calculate the following 1- √(〖(0-PF)〗^2+ 〖(1-PD)〗^2 )/√2

Aug 8, 2017

#1
+7827
+1

Expression

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

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

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

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

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

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

!  Thanks $$\large{X^2}$$

Aug 8, 2017
edited by asinus  Aug 8, 2017
#2
+2298
+2

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 $$(a-b)^2=a^2-2ab+b^2$$ $$(1-PD)^2$$ $$1^2-2(1)(PD)+(PD)^2$$ 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.
Aug 8, 2017