What is the value of x if (a + 2b – 3c)x^2 + (b + 2c – 3a)x + (c … - Vidyadip

MCQ

What is the value of $x$ if $(\text{a} + 2\text{b} – 3\text{c})\text{x}^2 + (\text{b} + 2\text{c} – 3\text{a})\text{x} + (\text{c} + 2\text{a} – 3\text{b}) = 0$ where $a, b, c$ are in $A.P?$

A
$\frac{1}{2}$
✓
$\frac{1}{4}$
C
$\frac{2}{3}$
D
$\frac{3}{4}$

✓

Answer

Correct option: B.

$\frac{1}{4}$

If the coefficients of $ (\text{x}^2 +\text{x} +\text{c}) = 0,$ then
$x$ will always be $= 1$
Therefore, here, $(\text{a} + 2\text{b} – 3\text{c}) + (\text{b} + 2\text{c} – 3\text{a}) + (\text{c} + 2\text{a} – 3\text{b}) = 0$
So, $x = 1.$
As, one of its root is $1$
so, we will calculate the other one.
As, $a, b, c$ are in $A.P$
so, $\text{b}=\Big(\frac{\text{a}+\text{c}}{2}\Big)$
Thus, product of the roots $\alpha\beta =\frac{(\text{c} + 2\text{a} – 3\text{b})}{(\text{a} + 2\text{b} – 3\text{c})}$
As, a root say $\alpha=1$ then,
$\beta=\frac{\Big(\frac{\text{c}+2\text{a}-3(\text{a}+\text{c})}{2}\Big)}{\Big(\frac{\text{a}+2(\text{a}+\text{c})}{2-3\text{c}}\Big)}$
We get the value of $\beta=\frac{1}{4}$

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