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Higher Order Equations

If both the r...

Question

If both the roots of $(2 a - 4) 9^{x} - (2 a - 3) 3^{x} + 1 = 0$ are non-negative, then

0 < a < 2

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2 < a <

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a <

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a > 3

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Solution

The correct option is B
2 < a <

Putting $3^{x}$ = y, we have

(2a - 4) $y^{2}$ - (2a - 3)y + 1 = 0

This equation must have real solution

$\Rightarrow$ $(2 a - 3)^{3}$ - 4 (2a - 4) $\geq$ 0

$\Rightarrow$ $4 a^{2}$ - 20a + 25 $\geq$ 0

$\Rightarrow$ $(2 a - 5)^{2} \geq 0$ . This is true.

y = 1 satisfies the equation

Since $3^{x}$ is positive and $3^{x} \geq 3^{\circ}, y \geq 1$

Product of the roots = 1 × y > 1 ×

$\Rightarrow \frac{1}{2 a - 4}$ > 1

$\Rightarrow$ 2a - 4 < 1 $\Rightarrow$ a < $\frac{5}{2}$

Sum of the roots $= \frac{2 a - 3}{2 a - 4}$ > 1

$\Rightarrow \frac{(2 a - 3) - (2 a - 4)}{2 a - 4}$ > 0

$\Rightarrow \frac{1}{2 a - 4}$ > 0

$\Rightarrow$ a > 2 $\Rightarrow$ 2 < a < $\frac{5}{2}$

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