In the real number system,the equation sqrt{x | vedclass

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(D) Given the equation: $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ Let $u = \sqrt{x-1}$,where $u \ge 0$. Then $x-1 = u^2$,so $x = u^2+1$. Sub

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infinitely many solutions

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(D) Given the equation: $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ Let $u = \sqrt{x-1}$,where $u \ge 0$. Then $x-1 = u^2$,so $x = u^2+1$. Substituting this into the equation: $\sqrt{u^2+1+3-4u} + \sqrt{u^2+1+8-6u} = 1$ $\sqrt{u^2-4u+4} + \sqrt{u^2-6u+9} = 1$ $\sqrt{(u-2)^2} + \sqrt{(u-3)^2} = 1$ $|u-2| + |u-3| = 1$ This equation holds if and only if $2 \le u \le 3$. Since $u = \sqrt{x-1}$,we have $2 \le \sqrt{x-1} \le 3$. Squaring the inequality: $4 \le x-1 \le 9$,which gives $5 \le x \le 10$. Thus,the equation has infinitely many solutions in the interval $[5, 10]$.

List-$I$	List-$II$
$(i)$ $\alpha = \beta$	$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$	$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$	$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$	$(D)$ $3b^2 = 16ac$
	$(E)$ $b^2 = 4ac$
	$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

In the real number system,the equation $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ has

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