Multiple choice

The number of real roots of the equation $(x+3)^{2}+(x+1)^{2}+(x-5)^{2}+(x-6)^{2}=0$ is

  1. $0$
  2. $2$
  3. $1$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A sum of squares equals zero if and only if each term is zero. Since (x+3)^2, (x+1)^2, (x-5)^2, and (x-6)^2 cannot be simultaneously zero for any real x, there are no real roots.

AI explanation

The equation is a sum of four squared terms equal to zero: (x+3)^2 + (x+1)^2 + (x-5)^2 + (x-6)^2 = 0. Since the square of any real number is always positive or zero, a sum of squares can only equal zero if each individual term equals zero simultaneously. For this to happen, x would need to be -3, -1, 5 and 6 at the same time, which is impossible. Therefore, there are 0 real roots.

Keep practicing — related questions