Question

Let f(x) = x2 + ax + b, where a, b > 0. 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, then find the value of ∛$b^a$ is

Answer 1

Answer:

2

Step-by-step explanation:

given

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

also,

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

substitute the values in the above equation

we get

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

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

by cancelling out both sides

we get

>> b = 2

given a>0

so in f(x) = x^2 + ax + b = 0

x should be -ve.(because f(x) = 0) ......( proof 1 )

we know

α + β = -a and αβ = b = 2

for αβ = 2 by (proof 1) both roots should be negative

hence α&β are negative

I. e

α = -1

β = -2

hence α + β = -1 + -2 = -3 = -a

hence, a = 3

$\sqrt[3]{ {b}^{a} } \: = \sqrt[3]{ {2}^{3} } = 2$

*************######**************

Answer 2

◇ QUESTION ◇:- Let f(x) = x2 + ax + b, where a, b > 0. 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, then find the value of ∛$b^a$ is?

◇ANSWER ◇:- 2.

☆SOLUTION ☆

GIVEN :-

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

also,

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

substituting the values in the above equation,

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

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

cancelling out both sides,

》 b = 2

》》 a>0 (given)

》so in f(x) = x^2 + ax + b = 0

》x should be -ve.(because f(x) = 0)

And,

》α + β = -a and αβ = b = 2

--> for αβ = 2 both roots should be negative

=> α&β are negative

》α = -1

》β = -2

》 α + β = -1 + -2 = -3 = -a

》hence, a = 3

☆ \sqrt[3]{ {b}^{a} } \: = \sqrt[3]{ {2}^{3} } = 23ba=323= 2 .

HOPE IT HELPS :)