If the roots of the equation $x^{3} - 12 x^{2} + 39 x - 28 = 0$ are in A.P., then their common difference will be $\pm 3$

True

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False

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Solution

The correct option is A True
Consider the given equation.
$x^{3} - 12 x^{2} + 39 x - 28 = 0$

Let the roots are $a - d, a, a + d$

So,
$a - d + a + a + d = 12$
$3 a = 12$
$a = 4$

And
$(a - d) (a) (a + d) = 28$
$a (a^{2} - d^{2}) = 28$
$4 (4^{2} - d^{2}) = 28$
$(16 - d^{2}) = 7$
$d^{2} = 9$
$d = \pm 3$

Hence, this is the answer.

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Similar questions

If the roots of the equation $x^{3} - 12 x^{2} + 39 x - 28 = 0$ are in A.P., then their common difference will be

x^{3} - 12 x^{2} + 39 x - 28 = 0

x^{3} - 12 x^{2} + 39 x - 28 = 0

x^{3} - 12 x^{2} + 39 x - 28 = 0

If the roots of the equation $x^{3} - 12 x^{2} + 39 x - 28 = 0$ are in A.P., then find the common difference.

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