If α and β be the zeroes of the polynomial 2x² + 3x – 6, find the values of

(i) α² + β² (ii) α² + β² + αβ

(iii) α²β + αβ² (iv)

(v) (vi) α – β

(vii) α³ + β³ (viii)

Let the quadratic polynomial be 2 x² + 3x – 6, and its zeroes are α and β.

We have

Here, a = 2 , b = 3 and c = – 6

….(1)

….(2)

(i) α² + β²

We have to find the value of α² + β²

Now, if we recall the identity

(a + b)² = a² + b² + 2ab

Using the identity, we get

(α + β)² = α² + β² + 2αβ

{from eqⁿ (1) & (2)}

(ii) α² + β² + αβ

{ from part (i)}

and, we have αβ = – 3

So,

=

=

(iii) α²β + α β²

Firstly, take common, we get

αβ(α + β)

and we already know the value of and .

So, α²β + α β² = αβ(α + β)

{from eqⁿ (1) and (2)}

(iv)

Let’s take the LCM first then we get,

(v)

Let’s take the LCM first then we get,

{from part(i) and eqⁿ (2)}

(vi)

Now, recall the identity

(a – b)² = a² + b² – 2ab

Using the identity , we get

(α – β)² = α² + β² – 2 αβ

{from part(i) and eqⁿ (2)}

= =

(vii)

Now, recall the identity

(a + b)³ = a³ + b³ + 3a² b + 3ab²

Using the identity, we get

⇒(α + β)³ = α³ + β³ + 3α² β + 3αβ²

(viii)

Let’s take the LCM first then we get,

{from part(vii) and eqⁿ (2)}