Question

if alpha and beta are zeroes of the polynomial 3x square +2x-6 then find the value of alpha cube +beta cube
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Answer 1

Answér :

α³ + ß³ = -116/27

Note:

★ The possible values of the variable for which the polynomial becomes zero are called its zeros .

★ A quadratic polynomial can have atmost two zeros .

★ The general form of a quadratic polynomial is given as ; ax² + bx + c .

★ If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then ;

• Sum of zeros , (α + ß) = -b/a

• Product of zeros , (αß) = c/a

★ If α and ß are the zeros of a quadratic polynomial , then that quadratic polynomial is given as : k•[ x² - (α + ß)x + αß ] , k ≠ 0.

★ The discriminant , D of the quadratic polynomial ax² + bx + c is given by ;

D = b² - 4ac

★ If D = 0 , then the zeros are real and equal .

★ If D > 0 , then the zeros are real and distinct .

★ If D < 0 , then the zeros are unreal (imaginary) .

Solution :

Here ,

The given quadratic polynomial is ;

3x² + 2x - 6 = 0 .

Comparing the given quadratic polynomial with the general quadratic polynomial ax² + bx + c , We have ;

a = 3

b = 2

c = -6

Also , α and ß be the zeros of the given quadratic polynomial .

Thus ,

Sum of zeros will be ;

=> α + ß = -b/a

=> α + ß = -2/3

Also ,

Product of zeros will be ;

=> αß = c/a

=> αß = -6/3

=> αß = -2

Now ,

=> (α + ß)³ = α³ + ß³ + 3αß•(α + ß)

=> (-2/3)³ = α³ + ß³ + 3(-2)(-2/3)

=> -8/27 = α³ + ß³ + 4

=> α³ + ß³ = -8/27 - 4

=> α³ + ß³ = (-8 - 108)/27

=> α³ + ß³ = -116/27

Hence ,

α³ + ß³ = -116/27