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.
+2668 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 \({{-4 - 2} \over 1 + 4 + 1 } = {-6 \over 6} = \color{brown}\boxed{-1}\)
+289 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.
