Question

solve this find roots of equation ​

Answer 1

Answer:

Step-by-step explanation:

Let f(x) = x3 -12x2 +39x -28 = 0

 

let us put x=1 , in above function

 

then f(1) = 1 -12 +39 -28 = 0.  Hence (x-1) is a factor of f(x)

 

when f(x) is divided by (x-1), we get quotient (x2 -11x +28) and this quotient is further factorised as (x-4)(x-7)

 

Hence f(x) = (x-1)(x-4)(x-7) = 0

 

Hence roots are 1, 4, 7 which are in arithmetic progression. first term is 1 and common difference is 3

                                                   OR

Let the roots of the given equation x^3−12x^2+39x−28=0 be a−d, a and a+d (Given that the roots are in A.P).

We know that the sum of the roots of a quadratic equation ax^3+bx^2+cx+d=0 is − b/a and the product of the roots is  

d/a

Here, the equation is x^3−12x^2+39x−28=0, therefore, we have:

Sum of the roots is:

(a−d)+a+(a+d)=−  (-12)/1

⇒3a=12

⇒a=  12/3

​  

 

⇒a=4.....(1)

Product of the roots is:

(a−d)a(a+d)=28

⇒(4−d)4(4+d)=28..................... ( From eqn (1) )

⇒(4−d)(4+d)=7

⇒4^2

−d^2  =7(∵x^2

−y ^2

=(x+y)(x−y))

⇒16−d^2  =7

⇒d ^2  =16−7

⇒d ^2  =9

⇒d=±  root of 9

​  

 

⇒d=±3  

Hence, the common difference is d=±3.

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Answer 2

Step-by-step explanation:

x=1 or 7 or 4

Hope this helps you