# 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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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.**

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.**