Find the roots of the equation: x^2 + x + 1 + (1/(x^2 + x + 1)) = ((10)/3)


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Question Number 149766 by mathdanisur last updated on 07/Aug/21

$Find the roots of the equation :$ $x^{2} + x + 1 + \frac{1}{x^{2} + x + 1} = \frac{10}{3}$
Commented by amin96 last updated on 07/Aug/21
$x^2 +x+1=y ⇒ y+(1/y)=((10)/3) y^2 +1=((10y)/3) ⇒ 3y^2 −10y+3=0 Δ=100−36=64 y=((10+8)/6)=3 x^2 +x+1=3 ⇒ x^2 +x−2=0 Δ=9 { ((x_1 =((−1+3)/2)=1)),((x_2 =((−1−3)/2)=−2)) :}$
$x^{2} + x + 1 = y \Rightarrow y + \frac{1}{y} = \frac{10}{3}$ $y^{2} + 1 = \frac{10 y}{3} \Rightarrow 3 y^{2} - 10 y + 3 = 0$ $Δ = 100 - 36 = 64 y = \frac{10 + 8}{6} = 3$ $x^{2} + x + 1 = 3 \Rightarrow x^{2} + x - 2 = 0 Δ = 9$ ${\begin{cases} x_{1} = \frac{- 1 + 3}{2} = 1 \\ x_{2} = \frac{- 1 - 3}{2} = - 2 \end{cases}$
Commented by mathdanisur last updated on 07/Aug/21

$Thankyou Ser$
	Answered by Canebulok last updated on 07/Aug/21

	$S o l u t i o n :$ $l e t :$ $\Rightarrow x^{2} + x + 1 = k$ $∵$ $\Rightarrow k + \frac{1}{k} = \frac{10}{3}$ $\Rightarrow k^{2} + 1 = \frac{10}{3} \cdot k$ $\Rightarrow k^{2} - \frac{10}{3} \cdot k + 1 = 0$ $B y u s i n g q u a d r a t i c f o r m u l a :$ $\Rightarrow k = \frac{(\frac{10}{3}) \pm \sqrt{{(- \frac{10}{3})}^{2} - 4}}{2}$ $= \frac{(\frac{10}{3}) \pm \sqrt{\frac{64}{9}}}{2}$ $∵$ $\Rightarrow k = 3; k = \frac{1}{3}$ $B y s u b s t i t u t i n g b a c k :$ $(1 s t s o l u t i o n s)$ $\Rightarrow x^{2} + x + 1 = 3$ $\Rightarrow x^{2} + x - 2 = 0$ $\Rightarrow (x + 2) (x - 1) = 0$ $\Rightarrow x_{1} = - 2; x_{2} = 1$ $(2 n d s o l u t i o n s)$ $\Rightarrow x^{2} + x + 1 = \frac{1}{3}$ $\Rightarrow x^{2} + x + \frac{2}{3} = 0$ $B y q u a d r a t i c f o r m u l a :$ $\Rightarrow x = \frac{- 1 \pm \sqrt{1^{2} - 4 (\frac{2}{3})}}{2}$ $x_{3, 4} = \frac{- 1 \pm i \sqrt{\frac{5}{3}}}{2}$ $S o l u t i o n b y : K e v i n$
	Commented by mathdanisur last updated on 07/Aug/21

	$Thankyou S er$
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