# 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)} Rate this question :

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