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

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 :

Accepted Answer

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

$=\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)}$

$\Rightarrow \sqrt{\mathrm{x}-3}=0 \Rightarrow \mathrm{x}=3$ which is in domain

or

$\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+3}=\sqrt{4 \mathrm{x}-2}$

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

$x^{2}+2 x-3=x^{2}-4 x+4$

$6 \mathrm{x}=7$

$\mathrm{x}=7 / 6$ (rejected)

Solutions · 1