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If $a,\;b,\;c$ are in Geometric Progression, then the equations $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$ have a common root if $\;\displaystyle\frac da,\;\frac eb,\;\frac fc$ are in:

  1. Arithmetic Progression
  2. Geometric Progression
  3. Harmonic Progression

Considering the first equation as $a_1x^2+b_1x+c_1=0$ and the second one as $a_2x^2+b_2x+c_2=0$, I applied the condition for the common root of two quadratic equations, i.e, $$(a_1b_2-b_1a_2)(b_1c_2-c_1b_2)=(c_1a_2-a_1c_2)^2$$ However, it gives a large equation in terms of the constants and does not lead me anywhere near finding the relation.

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6 Answers 6

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Hint: you haven't used the information that $a,b,c$ are in geometric progression. You can write $b=ar, c=ar^2$ and plug that into your condition, which simplifies it. You can also set $a=1$, which corresponds to dividing the original equation by $a$-if it is zero your equation is just $0=0$ You can plug the expression of each progression in for the second equation

If you continue to solve $x^2+rx+r^2=0$, you find the roots are proportional to $r$-so geometric progression clearly won't work as that says the two ratios are different.

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For two quadratic equations, $\begin{cases}a_0x^2+b_0x+c_0=0\\a_1x^2+b_1x+c_1=0\\\end{cases}$, if the following determinant
$\begin{vmatrix} a_0&b_0&c_0&0\\ 0&a_0&b_0&c_0\\ a_1&b_1&c_1&0\\ 0&a_1&b_1&c_1\\ \end{vmatrix}$
vanishes, then there is indeed a common root for the two.

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  • $\begingroup$ Could you please name the theorem on which this idea is based on? Thanks. $\endgroup$
    – NoChance
    Commented Mar 2, 2023 at 18:14
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HINT:

Let $$\frac cb=\frac ba=r\ne0\implies b=ar,c=ar^2$$

So, $ax^2+2bx+c=0\implies a(x^2+2rx+r^2)=0\implies x=-r$

$$\implies d(-r)^2+2e(-r)+f=0\implies r=\frac{e\pm\sqrt{e^2-df}}d$$

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  • $\begingroup$ A shorter approach would be to divide $dr^2-2er+f$ by $ar^2$. $\endgroup$ Commented Dec 9, 2015 at 9:31
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Do note that the discriminant of the first equation is $b^2-4ac$. Since a,b,c are in G.P, we must have $$b^2=ac$$ And we get the discriminant of the first quadratic to be 0. Hence $ax^2+2bx+c=0$ has equal roots. And since according to the question, $dx^2+2ex+f=0$ has a common root, it implies that both the quadratics are proportional to each other. $$dx^2+2ex+f=k(ax^2+2bx+c)$$ Finally comparing coefficients we get, $d=ak$,$2e=2bk$ and $f=ck$ and hence $$\frac da=\frac eb=\frac fc=k$$ this is the write ans

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Hint:

Let $a=r^2c,b=rc$. The first equation is

$$(r^2x^2+2rx+1)c=(rx+1)^2=0.$$

The root is double, $x=-\dfrac 1r$ !

Then

$$\frac d{r^2}-2\frac er+f=0,$$ and $$c\frac da-2c\frac eb+c\frac fc=0,$$ and we have an arithmetic progression.

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Using lab bhattarcharjee's method, $$d(-r)^2+2e(-r)+f=0$$ $$\therefore dr^2-2er+f=0$$Dividing throughout by $ar^2$, we get $$\frac da-\frac{2e}{ar}+\frac{f}{ar^2}=0$$$$\therefore \frac da+\frac{f}{c}=\frac {2e}{b}$$ Thus, $\displaystyle \frac da,\; \frac eb,\;\frac fc$ are in Arithmetic Progression.

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