Question

no useless answers........
solve the equation and find the roots 6x^4- 35x^3+62x^2-35x+6=0 ​

Answer 1

Answer:

2 , 3 , 1 / 2 , 1 / 3.

Step-by-step explanation:

Given,

 ⇒ 6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0

⇒ x^2( 6x^2 - 35x + 62 - 35/x + 6/x^2 ) = 0

  Now,

     Either x^2 is 0 or the remaining expression is 0. But x can't be 0, since f(x) will be 6, when x = 0.

Therefore,

⇒ 6x^2 - 35x + 62 - 35/x + 6/x^2 = 0

⇒ 6x^2 + 6/x^2 - 35x - 35/x + 62 = 0

⇒ 6( x^2 + 1 / x^2 ) - 35( x + 1 / x ) + 62 = 0

    ( x^2 + 1 /x^2 ) can also be written as ( x + 1 / x )^2 - 2( x * 1 / x ) or ( x + 1 / x )^2 - 2

 Continued :

⇒ 6[ ( x + 1 / x )^2 - 2 ] - 35( x + 1 / x ) + 62 = 0

   Let, x + 1 / x = a

⇒ 6( a^2 - 2 ) - 35a + 62 = 0

⇒ 6a^2 - 12 - 35a + 62 = 0

⇒ 6a^2 - 35a + 50 = 0

⇒ a = 5 / 2  or  10 / 3

   Therefore, case 1 :

⇒ x + 1 / x = 5 / 2

⇒ 2x^2 - 5x + 2 = 0

⇒ x = 1 / 2 or 2

   Case 2 :

⇒ x + 1 / x = 10 / 3

⇒ 3x^2 - 10x + 3 =0

⇒ x = 1 / 3 or 3

Answer 2

Answer:

See this.

Step-by-step explanation:

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