Question

if alpha and beta are the roots of equation x²-6x+1=0 than find the value of alpha⁴+beta⁴​

Answer 1

(Important note: For doing the question, alpha will be shown as @ and bita will be shown as ¶)

x²-6x+1

@+¶ = -b/a

@׶ = c/a

@+¶ = -(-6)/1

@+¶ = 6

@׶ = 1/1

@׶ = 1

@⁴+¶⁴ = (@²)²+(¶²)² [a²+b² = (a+b)²-2ab]

= (@²+¶²)²-2@¶ [(a²+b²)²-2ab = ((a+b)²-2ab)²-2ab]

= [(@+¶)²-2@¶]²-2@¶

On putting the values :-

= [(6)²-2×1]²-2×1

= [34]²-2

= 1156-2

@⁴+¶⁴ = 1154

(I hope the my answer would be understandable to you because the symbol replacing created a lot of problem to solve it)

Answer 2

Answer:

x²-6x +1

α+β=-b/a => 6

αβ= c/a =>1

α²+β²=(α+β)²−2αβ

(α²)²+(β²)²=(α²+β²)²-2α²β²

α⁴+β⁴=[((α+β)²-2αβ)²-2]

=34²-2

=1156-2=1154

Did you get the answer?