Question

solve this by quadratic equations​

Answer 1

Answer:

Factors : (x-2),(3x-9)

ROOTS : 2 , 3

Step-by-step explanation:

Let , x-1 = a

, x-2 = b

, x-3 = c

&, x-4 = d

So, Equation became ,

(1/ab) + (1/bc) + (1/cd) = 0

(bc²d + abcd + ab²c)/(ab²c²d)= 0

Dividing both sides by ab²c²d

bc²d + abcd + ab²c =0

bc (cd + ad + ab ) = 0

Dividing both sides by bc

cd + ad + ab =0

d(c+a) + ab =0

Now putting value of a , b , c , d

(x-4 )(x-3+x-1) + (x-1)(x-2) =0

(x-4)(2x-4) + (x-1)(x-2) =0

(x-4)(x-2)2 + (x-1)(x-2) =0

(x-2){2(x-4)+(x-1)} = 0

(x-2)(2x-8+x-1) =0

(x-1)(3x-9) =0

Hance Factors are : (x-1) , (3x-9)

Roots are :

x-1=0

x = 1

And

3x-9=0

3x=9

x = 9/3

x = 3

Roots ; 2 , 3

Hope you understand.

Answer 2

Answer:

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