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?

2 Answers
Jun 8, 2017

1

Explanation:

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

Jun 8, 2017

The answer is =1

Explanation:

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