If a,b and c are in G.P and the equations ax^2+2bx+c=0 and dx^2+2ex+f=0 have a common root,can you prove that d/a,e/b and f/c are in A.P?

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If a,b and c are in G.P and the equations ax^2+2bx+c=0 and dx^2+2ex+f=0 have a common root,can you prove that d/a,e/b and f/c are in A.P?

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Answer 1 · 2018-01-10T08:31:52

Please see below.

Explanation:

Consider $a x^{2} + 2 b x + c = 0$ $ax^2+2bx+c=0$

As $a, b$ $a,b$ and $c$ $c$ are in G.P., let $b = a r$ $b=ar$ and $c = a r^{2}$ $c=ar^2$

then the above becomes $a x^{2} + 2 a r x + a r^{2} = 0$ $ax^2+2arx+ar^2=0$

or $a (x^{2} + 2 r x + r^{2}) = 0$ $a(x^2+2rx+r^2)=0$ i.e. $a {(x + r)}^{2} = 0$ $a(x+r)^2=0$ and hence $x = - r$ $x=-r$ is the only root of $a x^{2} + 2 b x + c = 0$ $ax^2+2bx+c=0$ .

i.e. $- r$ $-r$ is also the root of ${d x}^{2} + 2 e x + f = 0$ $dx^2+2ex+f=0$

and we have $d r^{2} - 2 e r + f = 0$ $dr^2-2er+f=0$ and dividing this by $a r^{2}$ $ar^2$ we get

$\frac{d}{a} - \frac{2 e}{a r} + \frac{f}{a r^{2}} = 0$ $d/a-(2e)/(ar)+f/(ar^2)=0$

or d/a-(2e)/b+f/c=0#

or $\frac{d}{a} - \frac{e}{b} = \frac{e}{b} - \frac{f}{c}$ $d/a-e/b=e/b-f/c$

Hence, $\frac{d}{a}, \frac{e}{b}$ $d/a,e/b$ and $\frac{f}{c}$ $f/c$ are in A.P.

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If a,b and c are in G.P and the equations ax^2+2bx+c=0 and dx^2+2ex+f=0 have a common root,can you prove that d/a,e/b and f/c are in A.P?

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