solve-5-x-2-4x-2-3-x-2-4x-1-4-x-2-4x-8-

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

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