The number of real roots of the equation x 2 - 4 x + 3 + x 2 - 9 = 4 x 2 - 14 x + 6 , is [2023]

Question

The number of real roots of the equation $\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 6},$ is [2023]

Smart Achievers · Accepted Answer

(3)

$\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 6}$

$\Rightarrow \sqrt{(x - 3) (x - 1)} + \sqrt{(x - 3) (x + 3)} = \sqrt{2 (x - 3) (2 x - 1)}$

$\Rightarrow \sqrt{(x - 3)} [\sqrt{x - 1} + \sqrt{x + 3} - \sqrt{2} \sqrt{2 x - 1}] = 0$

$\Rightarrow x - 3 = 0 \Rightarrow x = 3$

$and \sqrt{x - 1} + \sqrt{x + 3} = \sqrt{2} \sqrt{2 x - 1}$

$\Rightarrow (x - 1) + (x + 3) + 2 \sqrt{(x - 1) (x + 3)} = 2 (2 x - 1)$

$\Rightarrow 2 x + 2 + 2 \sqrt{(x - 1) (x + 3)} = 4 x - 2$

$\Rightarrow x + 1 + \sqrt{(x - 1) (x + 3)} = 2 x - 1$

$\Rightarrow \sqrt{(x - 1) (x + 3)} = x - 2 \Rightarrow (x - 1) (x + 3) = x^{2} - 4 x + 4$

$\Rightarrow x^{2} + 2 x - 3 = x^{2} - 4 x + 4 \Rightarrow 6 x = 7 \Rightarrow x = \frac{7}{6}$

$Since, x^{2} - 4 x + 3 \geq 0 \Rightarrow (x - 3) (x - 1) \geq 0$

$\Rightarrow x \in (- \infty, 1] \cup [3, \infty)$ $...(i)$

$and x^{2} - 9 \geq 0 \Rightarrow (x - 3) (x + 3) \geq 0$

$\Rightarrow x \in (- \infty, - 3] \cup [3, \infty)$ $...(ii)$

$and 4 x^{2} - 14 x + 6 \geq 0 \Rightarrow (x - 3) (2 x - 1) \geq 0$

$\Rightarrow x \in (- \infty, \frac{1}{2}] \cup [3, \infty)$ $...(iii)$

$From (i), (ii), and (iii), x \in (- \infty, - 3] \cup [3, \infty)$

$Since x = \frac{7}{6} does not belong to x \in (- \infty, - 3] \cup [3, \infty) .$

$Therefore, x = 3 is the only solution.$

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