Solve-x-2-3x-2-x-2-8x-15-x-2-5x-6-x-2-10x-24-

Question Number 217888 by ArshadS last updated on 23/Mar/25

S o l v e :

\frac{x^{2} - 3 x + 2}{x^{2} - 8 x + 15} = \frac{x^{2} - 5 x + 6}{x^{2} - 10 x + 24}

Answered by bagjahebat last updated on 23/Mar/25

\frac{(x - 1) (x - 2)}{(x - 5) (x - 3)} = \frac{(x - 2) (x - 3)}{(x - 4) (x - 6)}

(x - 1) (x - 2) (x - 4) (x - 6) = (x - 2) (x - 3) (x - 5) (x - 3)

(x - 1) (x - 4) (x - 6) = {(x - 3)}^{2} (x - 5)

(x^{2} - 5 x + 4) (x - 6) = (x^{2} - 6 x + 9) (x - 5)

x^{3} - 6 x^{2} - 5 x^{2} + 30 x + 4 x - 24 = x^{3} - 5 x^{2} - 6 x^{2} + 30 x + 9 x - 45

x^{3} - 11 x^{2} + 34 x - 24 = x^{3} - 11 x^{2} + 39 x - 45

34 x - 24 = 39 x - 45

34 x - 39 x = - 45 + 24

- 5 x = - 21

x = \frac{21}{5}

Answered by Rasheed.Sindhi last updated on 23/Mar/25

\frac{(x - 1) (x - 2)}{(x - 3) (x - 5)} = \frac{(x - 2) (x - 3)}{(x - 4) (x - 6)}

(x - 1) (x - 2) (x - 4) (x - 6) - (x - 2) {(x - 3)}^{2} (x - 5) = 0

(x - 2) ((x - 1) (x - 4) (x - 6) - {(x - 3)}^{2} (x - 5)) = 0

x - 2 = 0 ∣ (x - 1) (x - 4) (x - 6) - {(x - 3)}^{2} (x - 5) = 0

x = 2

x^{3} - 11 x^{2} + 34 x - 24 - x^{3} + 11 x^{2} - 39 x + 45 = 0

- 5 x + 21 = 0 \Rightarrow x = \frac{21}{5}

Answered by Rasheed.Sindhi last updated on 23/Mar/25

A Tricky Method

\frac{x^{2} - 3 x + 2}{x^{2} - 8 x + 15} - 1 = \frac{x^{2} - 5 x + 6}{x^{2} - 10 x + 24} - 1

\frac{x^{2} - 3 x + 2}{x^{2} - 8 x + 15} - 1 = \frac{x^{2} - 5 x + 6}{x^{2} - 10 x + 24} - 1

\frac{5 x - 13}{x^{2} - 8 x + 15} = \frac{5 x - 18}{x^{2} - 10 x + 24}

\frac{x^{2} - 10 x + 24}{x^{2} - 8 x + 15} - 1 = \frac{5 x - 18}{5 x - 13} - 1

\frac{- 2 x + 9}{x^{2} - 8 x + 15} = \frac{- 5}{5 x - 13}

\frac{2 x - 9}{x^{2} - 8 x + 15} = \frac{5}{5 x - 13}

5 x^{2} - 40 x + 75 = 10 x^{2} - 71 x + 117

5 x^{2} - 31 x + 42 = 0

x = \frac{31 \pm \sqrt{961 - 840}}{10}

= \frac{31 \pm 11}{10} = \frac{21}{5}, 2