If the roots of the equation x^3-9x^2+23x-15=0,are in AP,then one of its roots will be (1)3,(2)9,(3)15(4)0

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justaguide's profile pic

Posted on

The roots of x^3 - 9x^2 + 23x - 15 = 0 are in AP.

x^3 - 9x^2 + 23x - 15 = 0

=> x^3 - 3x^2 - 6x^2 + 18x + 5x - 15 = 0

=> x^2(x -3 ) - 6x ( x - 3) + 5(x - 3) =0

=> (x^2 - 6x + 5)(x - 3) = 0

=> (x^2 - 5x - x + 5)(x - 3) = 0

=> (x(x - 5) - 1(x - 5))(x - 3) = 0

=> (x - 5)(x - 3)(x - 1) = 0

The roots are x = 1, x = 3 and x = 5. They are in AP.

The option (1) has a valid value of one of the roots, 3.

giorgiana1976's profile pic

Posted on

We'll impose the constraint that the roots of the given equation to be the terms of an AP:

x2 = (x1 + x3)/2

2x2 = x1 + x3

We'll apply Viete's relations and we'll have:

x1 + x2 + x3 = -(-9)/1

But x1 + x3 = 2x2

2x2 + x2 = 9

3x2 = 9

We'll divide by 3:

x2 = 3

The first option, (1)3, is the proper one, because one of the 3 roots of the equation is x = 3.

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