Question

if alpha and beta are zeroes of polynomial ax^2+bx+c.Find
1. alpha^3+beta^3
2. alpha-beta

Answer 1

α and β are the zeroes of the polynomial ax² + bx + c.

★ Sum of the zeroes : α + β = -b/a

★ Product of the zeroes : αβ = c/a

Case 1 :

α³ + β³ [ Given ]

α³ + β³ ⇒ ( α + β ) ( α² + β² - αβ )

α³ + β³ ⇒ ( α + β ) [(α² + β² +2αβ ) - 3αβ ]

α³ + β³ ⇒ ( α + β ) [ ( α + β )² - 3αβ ]

★ Putting the value

α³ + β³ ⇒ -b/a [ ( -b/a )² - 3c/a ]

α³ + β³ ⇒ -b/a [ b²/a² - 3c/a ]

α³ + β³ ⇒ -b/a ( b² -3ac / a² )

α³ + β³ ⇒ -b³ + 3ac / a³

Case 2 :

α - β [ Given ]

Squaring ,we get

( α - β )² ⇒ ( α + β )² - 4αβ

( α - β )² ⇒ ( α + β )² - 4αβ

★ Putting the values

( α - β )² ⇒ ( -b/a )² - 4 × c/a

( α - β )² ⇒ b²/a² - 4c/a

( α - β )² ⇒ b² - 4ac / a²

α - β ⇒ √b² - 4ac / a

Answer 2

SOLUTION :-

Let us assume that α and β are the zeroes of the polynomial ax² + bx + c.

As we know that,

• Sum of the zeroes : α + β = -b/a

• Product of the zeroes : α×β = c/a

In a first part,

α³ + β³ ------- [Given]

α³ + β³ => ( α + β ) ( α² + β² - αβ )

α³ + β³ => ( α + β ) [(α² + β² +2αβ ) - 3αβ ]

α³ + β³ => ( α + β ) [ ( α + β )² - 3αβ ]

•[Putting their values].

α³ + β³ => -b/a [ ( -b/a )² - 3c/a ]

α³ + β³ => -b/a [ b²/a² - 3c/a ]

α³ + β³ => -b/a ( b² -3ac / a² )

α³ + β³ => -b³ + 3ac / a³

In a second part,

α - β ------- [Given]

Squaring , (α - β) and we get,

( α - β )² => ( α + β )² - 4αβ

( α - β )² => ( α + β )² - 4αβ

•[Putting their values]

( α - β )² => ( -b/a )² - 4 × c/a

( α - β )² => b²/a² - 4c/a

( α - β )² => b² - 4ac / a²

α - β => √b² - 4ac / a

Answers :

• α³ + β³ = -b³ + 3ac / a³ .

• α - β = √b² - 4ac / a.