Q39 / 120 · NDA 2024 (Phase 2) · Mathematics

What is the number of real roots of the equation $(x-1)^2$ $+(x-3)^2+(x-5)^2=0 ?$

  • Correct answer
    A
    None
  • B
    Only one
  • C
    Only two
  • D
    Three

Solution

Given that, $(x-1)^2+(x-3)^2+(x-5)^2=0$ $\begin{aligned} & \Rightarrow x^2-2 x+1+x^2-6 x+9+x^2-10 x+25=0 \\ & \Rightarrow 3 x^2-18 x+35=0 \end{aligned}$ Discriminant $\mathrm{D}=(-18)^2-4 \times 3 \times 35=-96 < 0$ $\because$ Discriminant is negative. So, the given equation does not have any real roots.

Hence, the correct answer is option: 'A'.

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