JEE MainMathematicsEasyTheory of Equations

If $\alpha$ , $\beta$ and $\gamma$ are the roots of the equation
x^3 - 6x^2 + 11x - 6 = 0. Then, HM of the roots is

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We know that, if

\alpha

\beta

and

\gamma

are the roots
and A.G, H are AM, GM and HM of the roots, then equation is

x^{3} - 3Ax^{2} + \frac{3G^{3}}{H}x - G^{3} = 0

Given, x^3 - 6x^2+11x - 6 = 0
On comparing, we get 3A = 6...(i)

\frac{3G^{3}}{H} = 11

...(ii)
And G^3 = 6
From Equations (ii) and (iii), we get

H = \frac{18}{11}

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

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