Question

The number of real roots of the equation $(x-1{)}^2+(x-2{)}^2+(x-3{)}^2=0$ is :

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is D None of these

Given,

$(x-1{)}^2+(x-2{)}^2+(x-3{)}^2=0$

$\Rightarrow x^2-2x+1+x^2-4x+4+x^2-6x+9=0$

$\Rightarrow 3x^2-12x+14=0$

Let $D$ be the discriminant of the obtained quadratic equation.

$D=(-12{)}^2-4\times 14\times 3=144-168=-24$

Since the value of discriminant is negative, the equation does not have any real roots