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Find 1/(a - 1) + 1/(b - 1) where a and b are the roots of the quadratic equation 2x^2-7x+2 = x^2-11x+1.

 Jul 12, 2022
 #1
avatar+2448 
0

Simplify the equation to \(x^2 + 4x + 1 = 0\)

 

Note that\({1 \over (a - 1)} + {1 \over(b - 1)} = {(b-1) \over (a - 1)(b-1)} + {(a-1) \over(a-1)(b - 1)} = {{a + b - 2} \over ab−a−b+1} = {{(a + b )- 2} \over ab−(a+b)+1 }\)

 

Now, recall that \(a + b = -{b \over a} = -4\) and \(ab = {c \over a} = 1\)

 

Substituting this in gives us \({{-4 - 2} \over 1 + 4 + 1 } = {-6 \over 6} = \color{brown}\boxed{-1}\)

 Jul 12, 2022
 #2
avatar+279 
-2

Here is a different solution:

 

Root transformation!

 

2x^2 - 7x + 2 = x^2 - 11x + 1

x^2 + 4x - 1 = 0

 

If a, b are the solutions. Then a-1, b-1 are the solutions to the equation below:

 

(x+1)^2 + 4(x+1) - 1 = 0 

x^2 + 6x + 4 = 0

 

The reciprocal of the roots of this equation would be reversed:

4x^2 + 6x + 1 = 0.

 

The sum of the roots is -6 by vietas. 

 Jul 12, 2022

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