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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.

- 3 years, 8 months ago

Great!

- 3 years, 8 months ago

a p

- 3 years, 7 months ago

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