Question

Solve the following quadratic equation: (x-1) /(x+2)+(x-3)/(x-4)= 10/3

Answer 1

Given:

(x-1) /(x+2)+(x-3)/(x-4)= 10/3
(x-1)( x- 4) + (x-3)(x +2)/(x +2)(x-4) = 10/3
(x² -4x -x  + 4 + x² +2x -3x -6 ) / x²-4x +2x -8= 10/3

(x²  - 5x + 4 + x² - x -6)/ x²- 2x -8 = 10/3
(2x² -5x -x + 4 - 6)/x²-2x -8 = 10/3
2x² -6x - 2 /x²- 2x -8 = 10 /3
3 (2x² -6x -2) = 10(x²- 2x -8)
6x² -18x - 6 = 10x² -20x -80
6x² -10x² -18x +20x - 6 +80= 0
-4x² +2x + 74 = 0
-2(2x² - x - 37)= 0
2x² - x - 37= 0

Here, a= 2 , b= -1,c = - 37
x = -b ±√b² -4ac /2a  [quadratic formula]

x = -(-1)±√(-1)²- 4× 2 × -37) / 2× 2
x =( 1 ± √1 +296)/4
x = (1±√297)/4

HOPE THIS WILL HELP YOU...

Answer 2

hay!!

$\frac{x + 1}{x + 2} + \frac{x - 3}{x - 4} = \frac{10}{3} \\ \frac{(x - 1)(x - 4) + (x - 3)(x + 2)}{(x + 2)(x - 4)} = \frac{10}{3} \\ \frac{(x { }^{2} - 5x + 4 + {x}^{2} - x - 6)}{ {x}^{2} - 4x + 2x - 8} = \frac{10}{3} \\ \frac{( {x}^{2} - 5x + 4 + {x}^{2} - x - 6)}{ {x}^{2} - 2x - 8 } = \frac{10}{3} \\ \frac{2 {x}^{2} - 6x - 2}{ {x}^{2} - 2x - 8} = \frac{10}{3} \\ 3( 2{x}^{2} - 6x - 2) = 10( {x}^{2} - 2x - 8) \\ 6 {x}^{2} - 18x - 6 = 10 {x}^{2} - 20x - 80 = 0 \\ - 4 {x}^{2} + 2x + 74 = 0 \\ - 2 (2{x}^{2} - x - 37) = 0 \\ 2 {x}^{2} - x - 37 = 0 \\ \\ so \: \: \: a = 2 \: \: b = - 1 \: \: \: c = - 37$

then you want to value of

a, b, c so put it's
in shridharya charya fourmula
I hope it's help you

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