if (x^2 −(h+1)x+6i−6=0) then find the value of h if you know that one of the roots of the equation is three times the square of the other root (h∈ complex number)


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Question Number 114347 by mohammad17 last updated on 18/Sep/20

$i f (x^{2} - (h + 1) x + 6 i - 6 = 0) t h e n f i n d t h e v a l u e o f h$ $i f y o u k n o w t h a t o n e o f t h e r o o t s o f t h e e q u a t i o n i s t h r e e$ $t i m e s t h e s q u a r e o f t h e o t h e r r o o t$ $(h \in c o m p l e x n u m b e r)$
	Answered by Olaf last updated on 18/Sep/20

	$Two roots z_{1} and z_{2} with z_{1} = 3 z_{2}^{2}$ $z_{1} z_{2} = \frac{c}{a} = \frac{6 i - 6}{1} = - 3 \sqrt{2} e^{- i \frac{π}{4}} = 3 \sqrt{2} e^{i \frac{3 π}{4}}$ $3 z_{2}^{3} = 3 \sqrt{2} e^{i \frac{3 π}{4}}$ $z_{2}^{3} = \sqrt{2} e^{i \frac{3 π}{4}}$ $\Rightarrow z_{2} = 2^{\frac{1}{6}} e^{i \frac{π}{4}} or 2^{\frac{1}{6}} e^{i \frac{11 π}{12}} or 2^{\frac{1}{6}} e^{i \frac{19 π}{12}}$ $and z_{1} = 3 . 2^{\frac{1}{6}} e^{i \frac{π}{2}} or 3 . 2^{\frac{1}{6}} e^{i \frac{11}{6}} or 3 . 2^{\frac{1}{6}} e^{i \frac{19 π}{6}}$ $z_{1} + z_{2} = - \frac{b}{a} = h + 1$ $\Rightarrow h = z_{1} + z_{2} - 1$ $h = 2^{\frac{1}{6}} (3 e^{i \frac{π}{2}} + e^{i \frac{π}{4}} - 1) = 2^{\frac{1}{6}} (\frac{1}{\sqrt{2}} - 1 + i (\frac{1}{\sqrt{2}} + 3))$ $h = 2^{\frac{1}{6}} (3 e^{i \frac{11 π}{6}} + e^{i \frac{11 π}{12}} - 1)$ $h = 2^{\frac{1}{6}} (3 e^{i \frac{19 π}{6}} + e^{i \frac{19 π}{12}} - 1)$
	Commented by MJS_new last updated on 19/Sep/20
	$good method but −6+6i=6(√2)e^((3iπ)/4) 3z_2 ^3 =6(√2)e^((3iπ)/4) ⇒ z_2 = { ((1+i)),((w(1+i))),((w^2 (1+i))) :} with w=−(1/2)+((√3)/2)i$
	$good method but - 6 + 6 i = 6 \sqrt{2} e^{\frac{3 i π}{4}}$ $3 z_{2}^{3} = 6 \sqrt{2} e^{\frac{3 i π}{4}} \Rightarrow z_{2} = {\begin{cases} 1 + i \\ w (1 + i) \\ w^{2} (1 + i) \end{cases} with w = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$
	Commented by mohammad17 last updated on 19/Sep/20

	$t h a n k y o u s i r b u t n o w w h a t s t h e v a l u e o f (h)$ $c a n y o u s o l v e t h i s q u e s t i o n i n d e t a i l s p l e a s e ?$
	Commented by MJS_new last updated on 19/Sep/20
	$z_1 =3z_2 ^2 = { ((6i)),((6w^2 i)),((6w^4 i=6wi)) :} z_1 +z_2 =h+1 ⇔ h=z_1 +z_2 −1= { ((7i)),((−((3−5(√3))/2)−((7−(√3))/2)i)),((−((3+5(√3))/2)−((7+(√3))/2)i)) :}$
	$z_{1} = 3 z_{2}^{2} = {\begin{cases} 6 i \\ 6 w^{2} i \\ 6 w^{4} i = 6 w i \end{cases}$ $z_{1} + z_{2} = h + 1 \Leftrightarrow h = z_{1} + z_{2} - 1 = {\begin{cases} 7 i \\ - \frac{3 - 5 \sqrt{3}}{2} - \frac{7 - \sqrt{3}}{2} i \\ - \frac{3 + 5 \sqrt{3}}{2} - \frac{7 + \sqrt{3}}{2} i \end{cases}$
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