Question-124791

Question Number 124791 by ajfour last updated on 06/Dec/20

Commented by ajfour last updated on 06/Dec/20

I f p, q, r a r e r o o t s o f e q u a t i o n

x^{3} - x - c = 0, f i n d k .

Answered by ajfour last updated on 08/Dec/20

m = \tan θ = - \frac{r}{k}

\frac{2 m}{1 - m^{2}} = \tan 2 θ = \frac{p - r}{k}

{(- k m)}^{3} = (- k m) + c \dots . (I)

\frac{p}{k} = \frac{2 m - m + m^{3}}{1 - m^{2}} = \frac{m (m^{2} + 1)}{1 - m^{2}}

p = \frac{k m (m^{2} + 1)}{1 - m^{2}}

k^{3} m^{3} {(m^{2} + 1)}^{3} = {(1 - m^{2})}^{2} k m (m^{2} + 1)

+ c {(1 - m^{2})}^{3}

(k m - c) {(m^{2} + 1)}^{3} = {(1 - m^{2})}^{2} k m (1 + m^{2})

+ c {(1 - m^{2})}^{3}

4 {k m}^{3} (m^{2} + 1) = c {{(1 - m^{2})}^{3} + {(1 + m^{2})}^{3}}

\Rightarrow

4 {k m}^{3} (1 + m^{2}) = 2 c (1 + 3 m^{4}) \dots (I I)

f r o m (I) & (I I)

4 {k m}^{3} (1 + m^{2}) = 2 k m (1 - k^{2} m^{2}) (1 + 3 m^{4})

\Rightarrow 2 m^{2} (1 + m^{2}) = (1 - k^{2} m^{2}) (1 + 3 m^{4})

\Rightarrow 2 m^{2} + 2 m^{4} = 1 + 3 m^{4} - k^{2} m^{2} (1 + 3 m^{4})

\Rightarrow k^{2} m^{2} = \frac{m^{4} - 2 m^{2} + 1}{1 + 3 m^{4}} = \frac{{(1 - m^{2})}^{2}}{1 + 3 m^{4}}

& k^{2} m^{2} = \frac{c^{2} {(1 + 3 m^{4})}^{2}}{4 m^{4} {(1 + m^{2})}^{2}}

\Rightarrow \frac{{(1 - m^{2})}^{2}}{1 + 3 m^{4}} = \frac{c^{2} {(1 + 3 m^{4})}^{2}}{4 m^{4} {(1 + m^{2})}^{2}}

\Rightarrow \frac{4 m^{4} {(1 - m^{2})}^{2} {(1 + m^{2})}^{2}}{{(1 + 3 m^{4})}^{3}} = c^{2}

\Rightarrow \frac{4 t {(1 - t)}^{2}}{{(1 + 3 t)}^{3}} = c^{2}

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