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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