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Question Number 201081 by Calculusboy last updated on 29/Nov/23

	Answered by Rasheed.Sindhi last updated on 29/Nov/23

	$AnOther Way$ $2 x^{3} - x^{2} - 22 x - 24 = 0$ $l e t t h e t w o r o o t s a r e 3 a & 4 a w h e r e a \neq 0$ $∙ 2 {(3 a)}^{3} - {(3 a)}^{2} - 22 (3 a) - 24 = 0$ $54 a^{3} - 9 a^{2} - 66 a - 24 = 0 . . . . (i)$ $∙ 2 {(4 a)}^{3} - {(4 a)}^{2} - 22 (4 a) - 24 = 0$ $128 a^{3} - 16 a^{2} - 88 a - 24 = 0 . . . . (i i)$ $(i i) - (i) :$ $74 a^{3} - 7 a^{2} - 22 a = 0$ $74 a^{2} - 7 a - 22 = 0 [a \neq 0]$ $(2 a + 1) (37 a - 22) = 0$ $a = - \frac{1}{2}, \frac{22}{37} {(f a l s e)}^{★}$ $T w o r o o t s a r e 3 (- \frac{1}{2}) & 4 (- \frac{1}{2})$ $- \frac{3}{2} & - 2$ $l e t t h e t h i r d r o o t i s t$ $\overset{S u m o f t h e r o o t s}{- \frac{3}{2} - 2 + t = \frac{1}{2}} \land \overset{P r o d u c t o f t h e r o o t s}{(- \frac{3}{2}) (- 2) (t) = 12}$ $\Rightarrow t = 4$ $^{★} t w o r o o t s a r e 3 (\frac{22}{37}) & 4 (\frac{22}{37})$ $\frac{66}{37} & \frac{88}{37}$ $l e t t h e t h i r d r o o t i s t$ $\frac{66}{37} + \frac{88}{37} + t = \frac{1}{2} \land (\frac{66}{37}) (\frac{88}{37}) (t) = 12$ $t h a s n o v a l u e (s a t i s f y i n g b o t h)$ $∴ a = \frac{22}{37} i s f a l s e .$
	Commented by Calculusboy last updated on 29/Nov/23

	$t h a n k s s i r$
	Answered by Sutrisno last updated on 29/Nov/23

	$x_{1} = \frac{3}{4} x_{2}$ $x_{1} + x_{2} + x_{3} = \frac{1}{2}$ $\frac{3}{4} x_{2} + x_{2} + x_{3} = \frac{1}{2} \to x_{3} = \frac{1}{2} - \frac{7}{4} x_{2}$ $x_{1} . x_{2} . x_{3} = 12$ $\frac{3}{4} x_{3} . x_{2} (\frac{1}{2} - \frac{7}{4} x_{2}) = 12$ $21 x_{2}^{3} - 6 x_{2}^{2} + 192 = 0$ $(x_{2} + 2) (21 x_{2}^{2} - 48 x_{2} + 96) = 0$ $x_{2} = - 2$ $x_{1} = - \frac{3}{2}$ $x_{3} = 4$
	Commented by Calculusboy last updated on 29/Nov/23

	$t h a n k s s i r$
	Answered by mr W last updated on 29/Nov/23

	$x_{1} = 3 k$ $x_{2} = 4 k$ $x_{1} + x_{2} + x_{3} = \frac{1}{2}$ $\Rightarrow x_{3} = \frac{1}{2} - 7 k$ $x_{1} x_{2} + (x_{1} + x_{2}) x_{3} = - \frac{22}{2} = - 11$ $12 k^{2} + 7 k (\frac{1}{2} - 7 k) = - 11$ $74 k^{2} - 7 k - 22 = 0$ $\Rightarrow k = \frac{7 \pm 81}{148} = - \frac{1}{2}, \frac{22}{37}$ $x_{1} x_{2} x_{3} = 12 k^{2} (\frac{1}{2} - 7 k) = \frac{24}{2} = 12$ $\Rightarrow k^{2} (\frac{1}{2} - 7 k) = 1$ $o n l y k = - \frac{1}{2} i s o k$ $\Rightarrow x_{1} = - \frac{3}{2}, x_{2} = - 2, x_{3} = 4$
	Commented by Calculusboy last updated on 29/Nov/23

	$t h a n k s s i r$
	Answered by Rasheed.Sindhi last updated on 30/Nov/23
	$Let the roots are 3k,4k & λk { ((3k+4k+λk=1/2)),((3k.4k + 4k.λk + λk.3k = −11)),((3k.4k.λk=12)) :} ⇒ { ((k(7+λ)=1/2)),((k^2 (12 + 4λ + 3λ) = −11)),((k^3 (12λ)=12⇒λ=(1/k^3 ))) :} ⇒ { ((k(7+(1/k^3 ))=(1/2)⇒14k^3 −k^2 +2=0)),((k^2 (12 + 7((1/k^3 ))) = −11⇒12k^3 +11k+7=0)) :} ⇒ { (((2k+1)(7k^2 −4k+2)=0)),(((2k+1)(6k^2 −3k+7)=0)) :}⇒2k+1=0⇒k=−(1/2) ⇒λ=(1/k^3 )=(1/((−1/2)^3 ))=−8 Roots are: 3(−(1/2)) , 4(−(1/2)) , −8(−(1/2)) −(3/2) , −2 , 4$
	$L e t t h e r o o t s a r e 3 k, 4 k & λ k$ ${\begin{cases} 3 k + 4 k + λ k = 1 / 2 \\ 3 k . 4 k + 4 k . λ k + λ k . 3 k = - 11 \\ 3 k . 4 k . λ k = 12 \end{cases}$ $\Rightarrow {\begin{cases} k (7 + λ) = 1 / 2 \\ k^{2} (12 + 4 λ + 3 λ) = - 11 \\ k^{3} (12 λ) = 12 \Rightarrow λ = \frac{1}{k^{3}} \end{cases}$ $\Rightarrow {\begin{cases} k (7 + \frac{1}{k^{3}}) = \frac{1}{2} \Rightarrow 14 k^{3} - k^{2} + 2 = 0 \\ k^{2} (12 + 7 (\frac{1}{k^{3}})) = - 11 \Rightarrow 12 k^{3} + 11 k + 7 = 0 \end{cases}$ $\Rightarrow {\begin{cases} (2 k + 1) (7 k^{2} - 4 k + 2) = 0 \\ (2 k + 1) (6 k^{2} - 3 k + 7) = 0 \end{cases} \Rightarrow 2 k + 1 = 0 \Rightarrow k = - \frac{1}{2}$ $\Rightarrow λ = \frac{1}{k^{3}} = \frac{1}{{(- 1 / 2)}^{3}} = - 8$ $R o o t s a r e : 3 (- \frac{1}{2}), 4 (- \frac{1}{2}), - 8 (- \frac{1}{2})$ $- \frac{3}{2}, - 2, 4$
	Commented by Calculusboy last updated on 01/Dec/23

	$t h a n k s s i r$
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