Question

If $a,b,c$are in G.P, then the equations $a{x}^2+2bx+c=0$and $d{x}^2+2ex+f=0$ have a common root, if, $d/a,e/b,f/c$ are in

Answer 1

Step 1. Finding the terms $\mathit{d}\mathbf{/}\mathit{a}\mathbf{,}\mathbf{}\mathit{e}\mathbf{/}\mathit{b}\mathbf{,}\mathbf{}\mathit{f}\mathbf{/}\mathit{c}$ :

Given, $a,b,c$are in G.P

$b^2=ac$

$b=\sqrt{ab}...\left(i\right)$

Step 2. Put the value of equation (i) in equation $\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{0}\mathbf{}$

$a{x}^2+2\sqrt{ac}x+c=0$

$\Rightarrow$${(\sqrt{a}x+\sqrt{c})}^2=0$

$\Rightarrow$ $x=-\sqrt{(c/a)}$

Now,

$a{x}^2+2bx+c=0\&d{x}^2+2ex+f=0$have a common root

So $-\sqrt{c/a}$should satisfy $d{x}^2+2ex+f=0.$

Step 3. Put $\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\sqrt{\mathbf{c}\mathbf{/}\mathbf{a}}$ in $\mathit{d}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{e}\mathit{x}\mathbf{+}\mathit{f}\mathbf{=}\mathbf{0}$, we get

$d(c/a)-2e\sqrt{(c/a)}+f=0$

Divide it by $c$ ,

$(d/a)-2(e/\sqrt{ca})+(f/c)=0$

$\Rightarrow$ $(d/a)+(f/c)=2(e/b)$ ; $\left[\because b=\sqrt{ac}\right]$

Hence, the terms $\mathit{d}\mathbf{/}\mathit{a}\mathbf{,}\mathbf{}\mathit{e}\mathbf{/}\mathit{b}\mathbf{,}\mathbf{}\mathit{f}\mathbf{/}\mathit{c}$ are in A.P