Suppose \(a,b,c\) are three numbers in G.P. If the equations \(ax^2+2bx+c=0\) and \(dx^2+2ex+f\) have a common root, then \(\frac{d}{a} , \frac{e}{b} , \frac{f}{c}\) are in :

A.P.

G.P.

H.P.

none of the above.

**Note**: A.P.,G.P. and H.P. above indicate the arithmetic, geometric and harmonic progressions.

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TopNewestConsider two equations,

\(E_{1} : a_{1}x^2 + b_{1}x + c_{1}\)

\(E_{2} : a_{2}x^2 + b_{2}x + c_{2}\)

If they have a common root say \(\alpha\), then we can say that,

\(a_{1}\alpha^2 + b_{1}\alpha + c_{1} = 0\)

\(a_{2}\alpha^2 + b_{2}\alpha + c_{2} = 0\)

Solving them simultaneously for \(\alpha\) and \(\alpha^2\) and eliminating \(\alpha\) , we get an equation of the form,

\((a_{1}c_{2} - a_{2}c_{1})^{2} = (b_{1}c_{2} - b_{2}c_{1})(a_{1}b_{2} - a_{2}b_{1}) \)

Here, \(a_{1} = a, b_{1} = 2b, c_{1} = c, a_{2} = d, b_{2} = 2e, c_{2} = f \)

Substituting the values, we get,

\( (af - dc)^2 = 4(bf - ce)(ae - bd) \)

\((ac)^2(\frac{f}{c} - \frac{d}{a})^2 = 4ab^2c(\frac{f}{c} - \frac{e}{b})(\frac{e}{b} - \frac{d}{a}) \)

\( (\frac{f}{c} - \frac{d}{a})^2 = 4(\frac{f}{c} - \frac{e}{b})(\frac{e}{b} - \frac{d}{a}) \) (As \(a,b,c\) form a geometric progression)

Let \( \frac{d}{a} = l, \frac{e}{b} = m, \frac{f}{c} = n \). Therefore the above equation can be written as,

\((l - n)^2 + 4nl = 4mn + 4lm - 4m^2\)

\((l + n)^2 - 2\times2m(l+n) + (2m)^2 = 0\)

\((l + n -2m)^2 = 0\) which implies \(2m = l + n.\)

Therefore, they form an arithmetic progression. – Sudeep Salgia · 2 years, 11 months ago

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– Paramjit Singh · 2 years, 11 months ago

Great!Log in to reply

a p – Tejasvi Sharma · 2 years, 10 months ago

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