Question

In this question, x^y stands for x raised to the power y. For example, 2^3=8 and 4^1.5=8. If a,b are real numbers such that a+b=3, a^2+b^2=7, the value of a^4+b^4 is?

Answer 1

Answer:

a^4 + b^4 =  ((3 + √5 ) / 2)^4 +  ((3 - √5 ) / 2)^4

Step-by-step explanation:

we are given that

a + b = 3  ...(1)

a² + b² = 4  ....(2)

From equation (1) implies

a = 3 - b

Putting a = 3 - b in  equation (2) we get

    (3 - b)² + b² = 7

Using formula (a - b)² = a² + b² - 2ab we get

     9 + b² - 6b + b² = 7

⇒    2b² - 6b + 9 - 7 = 0

⇒     2b² - 6b + 2 = 0

Dividing by 2 on both sides we get

   b² - 3b + 1 = 0

Using quadric formula we get

   b = [ 3 + √(9 - 4) ] / 2 = ( 3 + √5) / 2

Or

   b = [ 3 - √(9 - 4) ] / 2 = ( 3 - √5) / 2

Now Putting the value of b = ( 3 - √5) / 2  in equation (1) we get

     a +  ( 3 - √5) / 2 = 3

⇒   a = 3 - ( 3 - √5) / 2

⇒   a = (3 + √5) / 2

Similarly when

b =  ( 3 + √5) / 2

a = (3 - √5 ) / 2

NOW

when

b =  ( 3 + √5) / 2

a = (3 - √5 ) / 2

a^4 + b^4 =  ((3 - √5 ) / 2)^4 +  ((3 + √5 ) / 2)^4

And when

b =  ( 3 - √5) / 2

a = (3 + √5 ) / 2

Then

a^4 + b^4 =  ((3 + √5 ) / 2)^4 +  ((3 - √5 ) / 2)^4