If $a,b,c$ are in Geometric Progression, then prove that the equations $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$ have a common root

Question

if $\frac da, \frac eb,\frac fc$ are in an Arithmetic Progression

$$b=\sqrt {ac}$$ $$ax^2+2\sqrt {ac} x +c=0$$ $$(\sqrt a x+\sqrt c)(\sqrt ax+\sqrt c)=0$$ $$x=-\sqrt{\frac ca}$$ since both roots are roots are equal, both equations have the same roots. Therefore $$\frac da =\frac eb =\frac fc$$

Why is this contradiction arising?

NOTE

As Martund correctly pointed out, I was wrong to assume both roots will satisfy the equation. Going by which $$d\frac ca -2e\sqrt{\frac ca} +f=0$$

How should I prove they are AP from here? I tried many combinations

Title says "geometric", text says "arithmetic". Which is it? — GEdgar, Dec 25 '19 at 12:24
@GEdgar I think the text must be the continuation of the question. — 19aksh, Dec 25 '19 at 12:33
@Narasimham, it is correctly stated, assuming that the text is a continuation of the title of the question. See my answer. — Martund, Dec 25 '19 at 12:42
Why should you prove the AP? It is one of the givens of the problem. — David K, Dec 25 '19 at 13:19
@DavidK, he wants to check whether it is an 'if and only if' statement. That is indeed true. Check my updated answer. — Martund, Dec 25 '19 at 13:23

Martund · Accepted Answer · 2019-12-25T13:20:34.230

This contradiction is arising because of the wrong statement that 'since both roots are roots are equal, both equations have the same roots'. Both roots of the first equation are same, does not imply that every quadratic equation having that root as one of the roots, has both roots equal.

Now proving the claim:

We wish to show that $x=-\sqrt\frac{c}{a}$ is a root of the equation $dx^2+2ex+f=0$ if $\frac{d}{a},\frac{e}{b}$ and $\frac{f}{c}$ are in AP. First let $r=\frac{e}{b}$ and $m=\frac{e}{b}-\frac{d}{a}=\frac{f}{c}-\frac{e}{b}$. Then applying quadratic formula in $dx^2+2ex+f=0$, we get, $$x = \frac{-e+\sqrt{e^2-df}}{d}$$ $$=\frac{-br+\sqrt{b^2r^2-ac(r-m)(r+m)}}{a(r-m)}$$ $$=\frac{-br+bm}{a(r-m)}$$ $$=-\sqrt\frac{c}{a}$$

Hence proved

EDIT Now, since the OP wants to prove that this is an 'if and only if' statement, here is the proof of the other direction. $$d\frac{c}{a}-2e\sqrt\frac{c}{a}+f=0$$ $$\iff dc-2eb+fa = 0$$ $$\iff dc+fa = 2eb$$ $$\iff \frac{d}{a}+\frac{f}{c}=2\frac{e}{b}$$ Last step follows by dividing both sides by $ac$. Also, note that all steps are iff in this proof, and hence it is proof of both directions, we need not do the earlier proof separately.

Yes that’s right, but can we prove it going backwards (check the edit — Aditya, Dec 25 '19 at 13:03

If $a,b,c$ are in Geometric Progression, then prove that the equations $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$ have a common root

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