If $a, b, c\in \mathbb{R}$ and equations $ ax^2 + bx + c= 0$ and $2x^2 + 4x + 6 = 0 $ have a common root with $a + b + c = 18$, the value of $a^2bc$ is
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Solution

Let $ax^2+bx+c = 0 $ ........................(1)

$ 2x^2+4x+6 = 0 $ ..........................(2)

Since $ D = b^2 - 4ac = 16 - 4 \times 2 \times 6 \\
~= - 32 (-ve) $

$\Rightarrow $Roots are imaginary

As the imaginary roots occurs in conjugate pairs. So, both the roots of the equation will be common.

So, $\dfrac{a}{2} = \dfrac{b}{4}= \dfrac{c}{6} $

$a : b : c = 1 : 2 : 3$

let $a = x$

$b = 2x$

$c = 3x$

$a + b + c = 18$

$x + 2x + 3x = 18$

$6x = 18\Rightarrow x = 3$

$\Rightarrow a = 3$

$\Rightarrow b = 6$

$\Rightarrow c = 9$

The value of $a^2bc = 3^2 \times 6 \times 9$

$= 9 \times 6 \times 9$

$~~~~~~= 486$

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