Question

If x and y are the zeroes of p(t)=t^2-5t+4 find 1/x - 1/y - 2x - y

Answer 1

$p(t) = {t}^{2} - 5t + 4 \\ = > {t}^{2} - 5t + 4 = 0 \\ = > {t}^{2} - 4t - t + 4 = 0 \\ = > t(t - 4) - 1(t - 4) = 0 \\ = > (t - 4)(t - 1) = 0$

either,

$t - 4 = 0 \\ = > t = 4$

or,

$t - 1 = 0 \\ = > t = 1$

X and Y are the zeroes of t

therefore,

first case

$x = 4 \\ y = 1$

$\frac{1}{x} - \frac{1}{y} - 2x - y \\ = \frac{1}{4} - \frac{1}{1} - 2 \times 4 - 1 \\ = \frac{1}{4} - 1 - 8 - 1 \\ = \frac{1}{4} - 10 \\ = \frac{1 - 40}{4} \\ = \frac{ - 39}{4}$

2nd case

$x = 1 \\ y = 4$

$\frac{1}{x} - \frac{1}{y} - 2x - y \\ = \frac{1}{1} - \frac{1}{4} - 2 \times 1 - 4 \\ = 1 - \frac{1}{4} - 2 - 4 \\ = - 5 - \frac{1}{4} \\ = \frac{ - 20 - 1}{4} \\ = \frac{ - 21}{4}$

so either the value is

$\frac{ - 39}{4} or \: \frac{ - 21}{4}$