Question

Let f(x)=x2+ax+b. if for all non zero real x f(x+1/x)=f(x)+f(1/x) and the roots of f(x)=0 are integers and what is the value of a2+b2​

Answer 1

Answer:

a² + b²  = 13

Step-by-step explanation:

f(x) = x² + ax + b

f(1/x) = 1/x² + a/x + b

f(x + 1/x)  =  (x + 1/x)² + a(x + 1/x) + b

= x² + 1/x² + 2  + ax + a/x + b

given that

f(x + 1/x) = f(x) + f(1/x)

=>  x² + 1/x² + 2  + ax + a/x + b  = x² + ax + b + 1/x² + a/x + b

=> 2= b

f(x)  = x² + ax  + 2

Product of roots  = 2

& roots are integer

=> Roots can be

-1 , - 2   ,  or  1  , 2

Sum of roots = - a

=> -a = -3  or 3

=> a = ±3

& b = 2

a² + b² = 9 + 4 = 13

a² + b²  = 13

Answer 2

Answer : (02)

f(x)= x^2+ax+b

f(x+1/x)= f(x) + f(1/x)

(x+1/x)^2 + a(x+1/x) + b = x^2+1/x^2+a(x+1/x)+2b

⇒ b=2

Since, x^2+ax+2 and roots are integers

⇒ a must also be an integer

For integral roots discriminant must be perfect square.

a^2-8=k^2 ⇒ (a-k)(a+k)= 8

Either,

a-k=1                           or              a-k=-1

a+k=8/2a=9                                 a+k=-8/2a=-9

a=9/2    (rejected)                          a=-9/2(rejected)

or a-k=-2                       or                 a-k=2

     a+k=-4/2a=-6                              a+k=4/2a=6

      a=-3                                              a=3

⇒ a=-3 (rejected)

∴∛2³ =∛8 = 2

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