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
Answers
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 : (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
Hey Mate,
Here’s your Answer
Please Mark it the Brainliest