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:

• A. 2
• B. 3
• C. 1
• D. 0

Answer 1

Answer: (C)

$ \sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)} $
$ =\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)} $
$ \Rightarrow \sqrt{x-3}=0 \Rightarrow x=3 \text { which is in domain }$
$ \text { or }$
$ \sqrt{x-1}+\sqrt{x+3}=\sqrt{4 x-2}$
$ 2 \sqrt{(x-1)(x+3)}=2 x-4$
$x^2+2 x-3=x^2-4 x+4 $
$ 6 x=7 $
$ x=7 / 6$(rejected)