Compute #1/[r-1] + 1/[s-1] + 1/[t-1]# given that #r,s# and #t# are the rootes of #x^3 -2x^2 +3x -4 = 0#?

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Answer 1 · 2017-06-08T16:42:39

#1#

# 1/[r-1] + 1/[s-1] + 1/[t-1] = (3-2(s+r+t)+rs +rt+st)/(rst-(rs+st+rt)+(r+s+t)-1) = (3-2(2)+3)/(4-3+2-1)=1#

Answer 2 · 2017-06-08T16:52:45

The answer is #=1#

If the roots of the equation

#x^3-2x^2+3x-4=0#

are #r#, #s# and #t#, we have

#r+s+t=-(-2)=2#

#rs+ts+rt=3#

#rst=-(-4)=4#

Therefore,

#1/(r-1)+1/(s-1)+1/(t-1)=((s-1)(t-1)+(r-1)(t-1)+(r-1)(s-1))/((r-1)(s-1)(t-1))#

The denominator is

#=(r-1)(s-1)(t-1)=(rs-r-s+1)(t-1)#

#=rst-rs-rt+r-st+s+t-1#

#=rst-(rs+rt+st)+(r+s+t)-1#

#=4-3+2-1=2#

The numerator is

#((s-1)(t-1)+(r-1)(t-1)+(r-1)(s-1))#

#=st-s-t+1+rt-r-t+1+rs-r-s+1#

#=(st+rt+rs)-2(r+s+t)+3#

#=3-2*2+3=2#

Therefore,

#1/(r-1)+1/(s-1)+1/(t-1)=2/2=1#