Question

find roots by quadratic formula (x-1/x+2)+(x-3/x-4) =10/3​

Answer 1

Answer:

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

L.C.M. of (x-2) & (x-4) = (x-2)(x-4)

=> [(x-1)(x-4)+(x-3)(x-2)]/(x-2)(x-4) = 10/3

=> 3(x^2 -5x +4 + x^2 -5x +6) = 10(x-2)(x-4)

=> 3( 2x^2–10x+10) = 10(x^2–6x+8)

=> 6x^2–30x +30 = 10x^2–60x+80

Taking 2 as a common from both sides & cancelling it, we have,

=> 3x^2–15x+15 = 5x^2–30x+40

=>5x^2–3x^2–30x+15x+40–15=0

=> 2x^2 -15x+25=0

Now, by the middle term splitting method,we have,

=> 2x^2 -10x-5x+25=0

=> 2x(x–5) -5(x-5) =0

=> (2x-5)(x-5)=0

=> 2x-5 = 0

So, x= 5/2

Or, x-5= 0

So, x=5

Answer 2

Solution:-

Equation :-

$\rm \to \bigg( \dfrac{x - 1}{x + 2} \bigg) + \bigg(\dfrac{x - 3}{x - 4} \bigg) = \dfrac{10}{3}$

Solving the equation

$\rm \to \dfrac{x - 1}{x + 2} + \dfrac{x - 3}{x - 4} = \dfrac{10}{3}$

Taking lcm

$\rm \to \: \dfrac{(x - 1)(x - 4) + (x - 3)(x + 2)}{(x + 2)(x - 4)} = \dfrac{10}{3}$

$\rm \: \to \dfrac{ {x}^{2} - 4x - x + </u></strong><strong><u>4</u></strong><strong><u> + {x}^{2} + 2x - 3x - 6 }{ {x}^{2} - 4x + 2x - 8} = \dfrac{10}{3}$

$\to \rm \: \dfrac{2 {x}^{2} - 5x + </u></strong><strong><u>4</u></strong><strong><u>- x - 6}{ {x}^{2} - 2x - 8} = \dfrac{10}{3}$

$\rm \to \: \dfrac{2 {x}^{2} - 6x </u></strong><strong><u>-</u></strong><strong><u> </u></strong><strong><u>2</u></strong><strong><u>}{ {x}^{2} - 2x - 8 } = \dfrac{10}{3}$

Using cross multiplication methods

$\rm \to \: 3(2 {x}^{2} - 6x </u></strong><strong><u>-</u></strong><strong><u> 2) = 10( {x}^{2} - 2x - 8)$

$\rm \to \: 6 {x}^{2} - 18x </u></strong><strong><u>-</u></strong><strong><u> 6 = 10 {x}^{2} - 20x - 80$

$\rm \to\: 10 {x}^{2} - 6 {x}^{2} - 20x + 18x - 80 </u></strong><strong><u>+</u></strong><strong><u> </u></strong><strong><u>6 = 0$

$\to \rm \: 4 {x}^{2} - 2x - </u></strong><strong><u>7</u></strong><strong><u>4</u></strong><strong><u> = 0$

$\rm \: \to \: 2(2 {x}^{2} - x - </u></strong><strong><u>3</u></strong><strong><u>7</u></strong><strong><u>) = 0$

$\rm \to2 {x}^{2} - x - </u></strong><strong><u>3</u></strong><strong><u>7</u></strong><strong><u> = 0$

Now find discriminant ( D)

Formula

$\boxed{ \rm \: D = {b}^{2} - 4ac}$

So

$\to \: \rm \: a \: = 2,b = - 1 \: and \: c = - 43$

putting the value

$\rm \: \to \: D = ( - 1) ^{2} - 4 \times 2 \times - </u></strong><strong><u>3</u></strong><strong><u>7</u></strong><strong><u>$

$\rm \to \: D = 1 + </u></strong><strong><u>2</u></strong><strong><u>9</u></strong><strong><u>6</u></strong><strong><u>$

$\to \rm \: D = </u></strong><strong><u>2</u></strong><strong><u>9</u></strong><strong><u>7</u></strong><strong><u>$

quadratic formula

$\rm \: x = \dfrac{ - b \pm \sqrt{D} }{2a}$

$\rm \: x = \dfrac{ - ( - 1) \pm \sqrt{</u></strong><strong><u>2</u></strong><strong><u>9</u></strong><strong><u>7</u></strong><strong><u>} }{2 \times 2}$

$\rm \: x = \dfrac{1 + </u></strong><strong><u>3</u></strong><strong><u> \sqrt{</u></strong><strong><u>3</u></strong><strong><u>3</u></strong><strong><u>} }{4} \: and \: x = \dfrac{1 - </u></strong><strong><u>3</u></strong><strong><u>\sqrt{</u></strong><strong><u>3</u></strong><strong><u>3</u></strong><strong><u>} }{4}$