If $a, b, c \in R$ and equations $a x^{2} + b x + c = 0$ and $2 x^{2} + 4 x + 6 = 0$ have a common root with $a + b + c = 18$ , the value of $a^{2} b c$ is
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486

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486.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

486.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

Let $a x^{2} + b x + c = 0$ ........................(1)

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

Since $D = b^{2} - 4 a c = 16 - 4 \times 2 \times 6 = - 32 (- v e)$

$\Rightarrow$ Roots are imaginary

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

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

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

let $a = x$

$b = 2 x$

$c = 3 x$

$a + b + c = 18$

$x + 2 x + 3 x = 18$

$6 x = 18 \Rightarrow x = 3$

$\Rightarrow a = 3$

$\Rightarrow b = 6$

$\Rightarrow c = 9$

The value of $a^{2} b c = 3^{2} \times 6 \times 9$

$= 9 \times 6 \times 9$

$= 486$

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