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.
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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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