QJEEM19Mafb8f: JEE Main 2019 Mathematics - Class 11 JEEM19Mafb8f Solution: If three distinct numbers a, b, c are in G.P. and the equations …

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

$\therefore$ b² = ac

In this equation ax² + 2bx + c = 0,

Discrimant, D = 4b² - 4ac

= 4ac - 4ac

= 0

Discrimant = 0 meand roots of the equation are equal.

Let both the roots of the equation = $\alpha$

$\therefore$ 2 $\alpha$ = $- {{2b} \over a}$

$\Rightarrow$ $\alpha$ = $- {b \over a}$

As both the equations ax² + 2bx + c = 0 and dx² + 2ex + ƒ = 0 have a common root,

so $- {b \over a}$ is also root of the equation dx² + 2ex + ƒ = 0.

$\therefore$ $- {b \over a}$ satisfy the equation dx² + 2ex + ƒ = 0.

$\therefore$ $d{\left( { - {b \over a}} \right)^2} + 2e\left( { - {b \over a}} \right) + f = 0$

$\Rightarrow$ $d{b^2} - 2aeb + f{a^2} = 0$

$\Rightarrow$ $dac - 2aeb + f{a^2} = 0$

$\Rightarrow$ $dc - 2eb + fa = 0$

$\Rightarrow$ ${{dc} \over {ac}} - {{2eb} \over {ac}} + {{fa} \over {ac}} = 0$

$\Rightarrow$ ${{dc} \over {ac}} - {{2eb} \over {{b^2}}} + {{fa} \over {ac}} = 0$

$\Rightarrow$ ${d \over a} - {{2e} \over b} + {f \over c} = 0$

$\Rightarrow$ ${{2e} \over b} = {d \over a} + {f \over c}$

$\therefore$ $d \over a$ , $e \over b$ , $f \over c$ are in A.P.