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So many P's!

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.

Note by Paramjit Singh
3 years, 8 months ago

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Consider 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 - 3 years, 8 months ago

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Great!

Paramjit Singh - 3 years, 8 months ago

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a p

Tejasvi Sharma - 3 years, 7 months ago

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