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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
	$m=tan θ=−(r/k) ((2m)/(1−m^2 ))=tan 2θ=((p−r)/k) (−km)^3 =(−km)+c ....(I) (p/k)=((2m−m+m^3 )/(1−m^2 ))=((m(m^2 +1))/(1−m^2 )) p=((km(m^2 +1))/(1−m^2 )) k^3 m^3 (m^2 +1)^3 =(1−m^2 )^2 km(m^2 +1) +c(1−m^2 )^3 (km−c)(m^2 +1)^3 =(1−m^2 )^2 km(1+m^2 ) +c(1−m^2 )^3 4km^3 (m^2 +1)=c{(1−m^2 )^3 +(1+m^2 )^3 } ⇒ 4km^3 (1+m^2 )=2c(1+3m^4 ) ...(II) from (I)&(II) 4km^3 (1+m^2 )=2km(1−k^2 m^2 )(1+3m^4 ) ⇒ 2m^2 (1+m^2 )=(1−k^2 m^2 )(1+3m^4 ) ⇒ 2m^2 +2m^4 =1+3m^4 −k^2 m^2 (1+3m^4 ) ⇒ k^2 m^2 =((m^4 −2m^2 +1)/(1+3m^4 )) = (((1−m^2 )^2 )/(1+3m^4 )) & k^2 m^2 =((c^2 (1+3m^4 )^2 )/(4m^4 (1+m^2 )^2 )) ⇒ (((1−m^2 )^2 )/(1+3m^4 ))=((c^2 (1+3m^4 )^2 )/(4m^4 (1+m^2 )^2 )) ⇒ ((4m^4 (1−m^2 )^2 (1+m^2 )^2 )/((1+3m^4 )^3 ))=c^2 ⇒ ((4t(1−t)^2 )/((1+3t)^3 ))=c^2 _________________________$
	$m = \tan θ = - \frac{r}{k}$ $\frac{2 m}{1 - m^{2}} = \tan 2 θ = \frac{p - r}{k}$ ${(- k m)}^{3} = (- k m) + c . . . . (I)$ $\frac{p}{k} = \frac{2 m - m + m^{3}}{1 - m^{2}} = \frac{m (m^{2} + 1)}{1 - m^{2}}$ $p = \frac{k m (m^{2} + 1)}{1 - m^{2}}$ $k^{3} m^{3} {(m^{2} + 1)}^{3} = {(1 - m^{2})}^{2} k m (m^{2} + 1)$ $+ c {(1 - m^{2})}^{3}$ $(k m - c) {(m^{2} + 1)}^{3} = {(1 - m^{2})}^{2} k m (1 + m^{2})$ $+ c {(1 - m^{2})}^{3}$ $4 {k m}^{3} (m^{2} + 1) = c {{(1 - m^{2})}^{3} + {(1 + m^{2})}^{3}}$ $\Rightarrow$ $4 {k m}^{3} (1 + m^{2}) = 2 c (1 + 3 m^{4}) . . . (I I)$ $f r o m (I) & (I I)$ $4 {k m}^{3} (1 + m^{2}) = 2 k m (1 - k^{2} m^{2}) (1 + 3 m^{4})$ $\Rightarrow 2 m^{2} (1 + m^{2}) = (1 - k^{2} m^{2}) (1 + 3 m^{4})$ $\Rightarrow 2 m^{2} + 2 m^{4} = 1 + 3 m^{4} - k^{2} m^{2} (1 + 3 m^{4})$ $\Rightarrow k^{2} m^{2} = \frac{m^{4} - 2 m^{2} + 1}{1 + 3 m^{4}} = \frac{{(1 - m^{2})}^{2}}{1 + 3 m^{4}}$ $& k^{2} m^{2} = \frac{c^{2} {(1 + 3 m^{4})}^{2}}{4 m^{4} {(1 + m^{2})}^{2}}$ $\Rightarrow \frac{{(1 - m^{2})}^{2}}{1 + 3 m^{4}} = \frac{c^{2} {(1 + 3 m^{4})}^{2}}{4 m^{4} {(1 + m^{2})}^{2}}$ $\Rightarrow \frac{4 m^{4} {(1 - m^{2})}^{2} {(1 + m^{2})}^{2}}{{(1 + 3 m^{4})}^{3}} = c^{2}$ $\Rightarrow \frac{4 t {(1 - t)}^{2}}{{(1 + 3 t)}^{3}} = c^{2}$ $_________________________$
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