Question

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

• A. $0$
• B. $2$
• C. $1$
• D. $Noneofthese$

Answer 1

The correct option is A $0$

Given that,

$(x+3{)}^2+(x+1{)}^2+(x-5{)}^2+(x-6{)}^2=0$

$⟹x^2+6x+9+x^2+2x+1+x^2-10x+25+x^2-12x+36=0$

$⟹4x^2-14x+71=0$

$\therefore b^2-4ac=(-14{)}^2-4(4\left)\right(71)=196-1136=-940<0$

Hence, given equation has no real roots

Option A is correct.