The number of real roots of the equation sqrt | vedclass

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(B) The given equation is $\sqrt{(x-1)(x-3)} + \sqrt{(x-3)(x+3)} = \sqrt{(x-3)(4x-2)}$.Case $1$: $\sqrt{x-3} = 0 \implies x = 3$.Checking the domain:

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$1$

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(B) The given equation is $\sqrt{(x-1)(x-3)} + \sqrt{(x-3)(x+3)} = \sqrt{(x-3)(4x-2)}$.Case $1$: $\sqrt{x-3} = 0 \implies x = 3$.Checking the domain: For $\sqrt{x^2-9}$,we need $x^2-9 \ge 0$,so $x \ge 3$ or $x \le -3$. For $x=3$,$\sqrt{0} + \sqrt{0} = \sqrt{0}$,which is true. Thus,$x=3$ is a root.Case $2$: $\sqrt{x-3} 
eq 0$. Dividing by $\sqrt{x-3}$ (assuming $x > 3$):$\sqrt{x-1} + \sqrt{x+3} = \sqrt{4x-2}$.Squaring both sides:$(x-1) + (x+3) + 2\sqrt{(x-1)(x+3)} = 4x-2$.$2x + 2 + 2\sqrt{x^2+2x-3} = 4x-2$.$2\sqrt{x^2+2x-3} = 2x-4$.$\sqrt{x^2+2x-3} = x-2$.Squaring again:$x^2+2x-3 = x^2-4x+4$.$6x = 7 \implies x = 7/6$.Since $x=7/6$ does not satisfy $x \ge 3$,it is rejected.Therefore,there is only $1$ real root.

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

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