4x^4 +16x^3 +24x^2 −9x−1=0 using any method. find real value of x that satisfy the polynomial


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Question Number 41606 by psyche-ace last updated on 10/Aug/18

$4 x^{4} + 16 x^{3} + 24 x^{2} - 9 x - 1 = 0$ $using any method . find real value of x that satisfy the polynomial$
	Answered by MJS last updated on 10/Aug/18

	$f (x) = x^{4} + 4 x^{3} + 6 x^{2} - \frac{9}{4} x - \frac{1}{4} = 0$ $f^{'} (x) = 4 x^{3} + 12 x^{2} + 12 x - \frac{9}{4}$ $x^{3} + 3 x^{2} + 3 x - \frac{9}{16} = 0$ $x = t - 1$ $t^{3} - \frac{25}{16} = 0 \Rightarrow t = \frac{\sqrt[3]{100}}{4} \Rightarrow x = \frac{- 4 + \sqrt[3]{100}}{4}$ $f^{'} (x) has got only 1 real root \Rightarrow f (x) has 0 or 2 real roots$ $f^{″} (x) = 12 x^{2} + 24 x + 12$ $f (\frac{- 4 + \sqrt[3]{100}}{4}) = \frac{5}{64} (64 - 15 \sqrt[3]{100}) \approx - . 44$ $f^{″} (\frac{- 4 + \sqrt[3]{100}}{4}) = \frac{15 \sqrt[3]{10}}{2} > 0 \Rightarrow f (x) has 2 real roots$ $let$ $f (x) = (x - a - \sqrt{b}) (x - a + \sqrt{b}) (x - c - \sqrt{d} i) (x - c + \sqrt{d} i)$ $\Rightarrow$ $a + c + 2 = 0$ $a^{2} + 4 a c - b + c^{2} + d - 6 = 0$ $8 a^{2} c + 8 {a c}^{2} + 8 a d - 8 b c - 9 = 0$ $a^{2} c^{2} + a^{2} d - {b c}^{2} - b d + \frac{1}{4} = 0$ $a = - c - 2$ $- b - 2 c^{2} - 4 c + d - 2 = 0$ $- 8 b c + 16 c^{2} - 8 c d + 32 c - 16 d - 9 = 0$ $- {b c}^{2} - b d + c^{4} + 4 c^{3} + c^{2} d + 4 c^{2} + 4 c d + 4 d + \frac{1}{4} = 0$ $a = - c - 2$ $b = - 2 c^{2} - 4 c + d - 2$ $16 c^{3} + 48 c^{2} - 16 c d + 48 c - 16 d - 9 = 0$ $3 c^{4} + 8 c^{3} + 2 c^{2} d + 6 c^{2} + 8 c d - d^{2} + 6 d + \frac{1}{4} = 0$ $a = - c - 2$ $b = - 2 c^{2} - 4 c + d - 2$ $d = \frac{16 c^{3} + 48 c^{2} + 48 c - 9}{16 (c + 1)}$ $4 c^{4} + 16 c^{3} + 24 c^{2} + 16 c - 1 - \frac{625}{256 {(c + 1)}^{2}} = 0$ $c^{6} + 6 c^{5} + 15 c^{4} + 20 c^{3} + \frac{55}{4} c^{2} + \frac{7}{2} c - \frac{881}{1024} = 0$ $c = t - 1$ $t^{6} - \frac{5}{4} t^{2} - \frac{625}{1024} = 0$ $t = \sqrt{z}$ $z^{3} - \frac{5}{4} z - \frac{625}{1024} = 0$ $z = \frac{1}{48} (\sqrt[3]{30 (1125 + \sqrt{282585}} + \sqrt[3]{30 (1125 - \sqrt{282585}}) \approx$ $\approx 1.30994$ $t \approx 1.14453$ $c \approx . 144525$ $a = - 2.14453$ $b = - 2.67513$ $d = - . 055256$ $a \pm \sqrt{b} = - 2.14453 \pm 1 . 63558 i$ $c + \sqrt{d} i = - . 0905405$ $c - \sqrt{d} i = . 379591$
	Commented by MJS last updated on 10/Aug/18

	$to eliminate the {b x}^{n - 1} you can always set$ $x = t - \frac{b}{n} if the constant factor if {a x}^{n} is equal$ $to 1 (a = 1)$ $examples :$ $3 x^{2} + 4 x - 5 = 0$ $x^{2} + \frac{4}{3} x - \frac{5}{3} = 0$ $x = t - \frac{2}{3}$ ${(t - \frac{2}{3})}^{2} + \frac{4}{3} (t - \frac{2}{3}) - \frac{5}{3} = 0$ $t^{2} - \frac{19}{9} = 0$ $(which in this case is the same as the usual$ $formula x = - \frac{p}{2} \pm \sqrt{\frac{p^{2}}{4} - q}$ $4 x^{3} - 5 x^{2} + 3 x - 2 = 0$ $x^{3} - \frac{5}{4} x^{2} + \frac{3}{4} x - \frac{1}{2} = 0$ $x = t + \frac{5}{12}$ ${(t + \frac{5}{12})}^{3} - \frac{5}{4} {(t + \frac{5}{12})}^{2} + \frac{3}{4} (t + \frac{5}{12}) - \frac{1}{2} = 0$ $t^{3} + \frac{11}{48} t - \frac{287}{864} = 0$ $(which in this case leads to {Cardano}^{'} s formula$ $or in case of 3 real solutions to the trigonometric$ $method)$ $in case of a 4^{th} - degree polynome with 2 real$ $and 2 complex roots my method leads to a$ $polynome of 6^{th} degree which can be reduced$ $to one of 3^{rd} degree . . .$ $will post an example later$
	Commented by Tawa1 last updated on 10/Aug/18

	$Sir how can we know that x = t - 1$
	Commented by Tawa1 last updated on 10/Aug/18

	$wow, please send it sir . God bless you sir$
	Commented by Tawa1 last updated on 10/Aug/18

	$Am still expecting the other method and examples sir . God bless you$
	Commented by Tawa1 last updated on 10/Aug/18

	$i will love to see the formulars sir$
	Commented by Tawa1 last updated on 11/Aug/18

	$Sir MJS am still expecting sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18
	$4x^2 (x^2 +4x+4)+8x^2 −9x−1=0 {(2x)(x+2)}^2 +9x^2 −6x+1−x^2 −3x−2=0 {(2x)(x+2)}^2 +(3x−1)^2 −(x^2 +3x+2)=0 {(2x)(x+2)}^2 +(3x−1)^2 −{x^2 +2.x.(3/2)+(9/4)+2−(9/4)}=0 {(2x)(x+2)}^2 +(3x−1)^2 −{(x+(3/2))^2 −(1/4)}=0 [{(2x)(x+2)}^2 +(3x−1)^2 +(1/4)]−[(x+(3/2))^2 ]=0 now i want to say some thing the value of [{(2x)(x+2)}^2 +(3x−1)^2 +(1/4)] is +ve and say its value is p now the value of [(x+(3/2))^2 ] is also positve and say value is q now the eqn is p−q=0 that imply the value of p should be equal to q to make the equation true but {(2x)(x+2)}^2 >[(x+(3/2))^2 ] so to me p can not be equal to q to make the equation true.. to me p>>q so i think this equation is not feasible...$
	$4 x^{2} (x^{2} + 4 x + 4) + 8 x^{2} - 9 x - 1 = 0$ ${(2 x) (x + 2)}^{2} + 9 x^{2} - 6 x + 1 - x^{2} - 3 x - 2 = 0$ ${(2 x) (x + 2)}^{2} + {(3 x - 1)}^{2} - (x^{2} + 3 x + 2) = 0$ ${(2 x) (x + 2)}^{2} + {(3 x - 1)}^{2} - {x^{2} + 2 . x . \frac{3}{2} + \frac{9}{4} + 2 - \frac{9}{4}} = 0$ ${(2 x) (x + 2)}^{2} + {(3 x - 1)}^{2} - {{(x + \frac{3}{2})}^{2} - \frac{1}{4}} = 0$ $[{(2 x) (x + 2)}^{2} + {(3 x - 1)}^{2} + \frac{1}{4}] - [{(x + \frac{3}{2})}^{2}] = 0$ $n o w i w a n t t o s a y s o m e t h i n g$ $t h e v a l u e o f$ $[{(2 x) (x + 2)}^{2} + {(3 x - 1)}^{2} + \frac{1}{4}] i s + v e a n d s a y i t s$ $v a l u e i s p$ $n o w t h e v a l u e o f [{(x + \frac{3}{2})}^{2}] i s a l s o p o s i t v e a n d$ $s a y v a l u e i s q$ $n o w t h e e q n i s p - q = 0$ $t h a t i m p l y t h e v a l u e o f p s h o u l d b e e q u a l t o q t o$ $m a k e t h e e q u a t i o n t r u e$ $b u t {(2 x) (x + 2)}^{2} > [{(x + \frac{3}{2})}^{2}]$ $s o t o m e p c a n n o t b e e q u a l t o q t o m a k e$ $t h e e q u a t i o n t r u e . .$ $t o m e p >> q$ $s o i t h i n k t h i s e q u a t i o n i s n o t f e a s i b l e . . .$
	Commented by Tawa1 last updated on 10/Aug/18

	$God bless you sir$
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