solve (5/(x^2 +4x+2))=(3/(x^2 +4x+1))−(4/(x^2 +4x+8))


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Question Number 172027 by Mikenice last updated on 23/Jun/22

$s o l v e$ $\frac{5}{x^{2} + 4 x + 2} = \frac{3}{x^{2} + 4 x + 1} - \frac{4}{x^{2} + 4 x + 8}$
Commented by Mikenice last updated on 23/Jun/22

$t h a n k s s i r$
	Answered by Rasheed.Sindhi last updated on 24/Jun/22

	$L e t x^{2} + 4 x + 1 = y$ $\frac{5}{y + 1} = \frac{3}{y} - \frac{4}{y + 7}$ $5 y (y + 7) = 3 (y + 1) (y + 7) - 4 y (y + 1)$ $5 y^{2} + 35 y = 3 y^{2} + 24 y + 21 - 4 y^{2} - 4 y$ $6 y^{2} + 15 y - 21 = 0$ $2 y^{2} + 5 y - 7 = 0$ $(y - 1) (2 y + 7) = 0$ $y = 1 ∣ y = - \frac{7}{2}$ $x^{2} + 4 x + 1 = 1 ∣ x^{2} + 4 x + 1 = - \frac{7}{2}$ $x (x + 4) = 0 ∣ 2 x^{2} + 8 x + 9 = 0$ $(x = 0 ∣ x = - 4) ∣ x = \frac{- 8 \pm \sqrt{64 - 72}}{4}$ $(x = 0 ∣ x = - 4) ∣ x = \frac{- 4 \pm \sqrt{2}}{2}$
	Commented by Mikenice last updated on 23/Jun/22

	$t h a n k s s i r$
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