If a, b, c are three real roots of the equation x3−6x2+5x−1=0x3-6x2+5x-1=0, then a4+b4+c4a4+b4+c4 is equal
Answers
Correct Question:
If a, b, c are three real roots of the equation x³ - 6x² + 5x - 1 = 0, then find the value of a⁴ + b⁴ + c⁴.
Answer:
650
Step-by-step explanation:
Given:
a, b, c are the real roots of the equation x³ - 6x² + 5x - 1 = 0
Comparing the given equation with a'x³ + b'x² + c'x + d = 0 we get,
- a' = 1
- b' = - 6
- c' = 5
- d = - 1
Sum of roots = a + b + c = - b'/a' = - ( - 5 )/1 = 5
Product of roots = abc = - d/a' = - ( - 1 )/1 = = 1
Sum of product of roots taken two at a time = ab + bc + ca = c'/a' = 5/1 = 5
First let's find the value of a² + b² + c²
a + b + c = 6
Squaring on both sides
⇒ ( a + b + c )² = 6²
Since ( a + b + c )² = a² + b² + c² + 2( ab + bc + ca )
⇒ a² + b² + c² + 2( ab + bc + ca ) =36
⇒ a² + b² + c² + 2( 5 ) = 36
⇒ a² + b² + c² + 10 = 36
⇒ a² + b² + c² = 36 - 10
⇒ a² + b² + c² = 26
Squaring on both sides
⇒ ( a² + b² + c² )² = 26²
Since ( x + y + z )² = x² + y² + z² + 2( xy + yz + zx )
⇒ ( a² )² + ( b² )² + ( c² )² + 2( a²b² + b²c² + c²a² ) = 676
⇒ a⁴ + b⁴ + c⁴ + 2( a²b² + b²c² + c²a² ) = 676 → ( 1 )
To find the value of a⁴ + b⁴ + c⁴ we need to find the value of a²b² + b²c² + c²a²
ab + bc+ ca = 5
Squaring on both sides
⇒ ( ab + bc + ca )² = 5²
Since ( x + y + z )² = x² + y² + z² + 2( xy + yz + zx )
⇒ ( ab )² + ( bc )² + ( ca )² + 2{ ab(bc) + bc(ca) + ca(ab) } = 25
⇒ a²b² + b²c² + c²a² + 2{ abc( b + c + a ) } = 25
⇒ a²b² + b²c² + c²a² + 2{ 1( 6 ) } = 25
⇒ a²b² + b²c² + c²a² + 12 = 25
⇒ a²b² + b²c² + c²a² = 25 - 12
⇒ a²b² + b²c² + c²a² = 13
Substituting the value of a²b² + b²c² + c²a² in ( 1 )
⇒ a⁴ + b⁴ + c⁴ + 2( a²b² + b²c² + c²a² ) = 676
⇒ a⁴ + b⁴ + c⁴ + 2( 13 ) = 676
⇒ a⁴ + b⁴ + c⁴ + 26 = 676
⇒ a⁴ + b⁴ + c⁴ = 676 - 26
⇒ a⁴ + b⁴ + c⁴ = 650