Question

The number of roots of the equation $\sqrt{x-2}(x^2-4x+3)=0$ is/are:

• A. three
• B. four
• C. one
• D. two

Answer 1

Answer: (D)

two

From the above expression, we get$\sqrt{x-2}=0$ Or $x^2-4x+3=0$
$\sqrt{x-2}=0$ gives $x=2$
And $x^2-4x+3=0$$\Rightarrow (x-2{)}^2-4+3=0$$\Rightarrow (x-2{)}^2=1$$\Rightarrow (x-2)=\pm 1$$\Rightarrow x=3$ or $x=1$
However $x=1$ is not applicable since domain of $\sqrt{x-2}$ is $x>2$
The roots are therefore $2$ and $3$.
Hence, the number of roots will be 2.