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.

#Goldbach'sConjurersGroup #ArithmeticGeometricProgression

Note by A Brilliant Member
7 years, 3 months ago

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Comments

Consider two equations,
E1:a1x2+b1x+c1E_{1} : a_{1}x^2 + b_{1}x + c_{1}
E2:a2x2+b2x+c2E_{2} : a_{2}x^2 + b_{2}x + c_{2}
If they have a common root say α\alpha, then we can say that,
a1α2+b1α+c1=0a_{1}\alpha^2 + b_{1}\alpha + c_{1} = 0
a2α2+b2α+c2=0a_{2}\alpha^2 + b_{2}\alpha + c_{2} = 0
Solving them simultaneously for α\alpha and α2\alpha^2 and eliminating α\alpha , we get an equation of the form,
(a1c2−a2c1)2=(b1c2−b2c1)(a1b2−a2b1)(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, a1=a,b1=2b,c1=c,a2=d,b2=2e,c2=fa_{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) (af - dc)^2 = 4(bf - ce)(ae - bd)
(ac)2(fc−da)2=4ab2c(fc−eb)(eb−da)(ac)^2(\frac{f}{c} - \frac{d}{a})^2 = 4ab^2c(\frac{f}{c} - \frac{e}{b})(\frac{e}{b} - \frac{d}{a})

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

Let da=l,eb=m,fc=n \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−4m2(l - n)^2 + 4nl = 4mn + 4lm - 4m^2
(l+n)2−2×2m(l+n)+(2m)2=0(l + n)^2 - 2\times2m(l+n) + (2m)^2 = 0
(l+n−2m)2=0(l + n -2m)^2 = 0 which implies 2m=l+n.2m = l + n.
Therefore, they form an arithmetic progression.

Sudeep Salgia - 7 years, 3 months ago

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Great!

A Brilliant Member - 7 years, 3 months ago

a p

Tejasvi Sharma - 7 years, 3 months ago
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