# 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)}

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