Question

Write number of real roots of the equation (x+1)^2+(x-2)^2+(x-3)^2=0

Answer 1

Answer:

Hello!

Step by step explanation is given below-

Step-by-step explanation:

This question is very easy if you know the following property of perfect squares -

"If the sum of any n number of perfect squares equals to 0 , then each square equals to 0 as well"

So according to this property -

(x+1)^2 =0----------(equation number 1)

(x-2)^2=0-----------(equation number 2)and

(x-3)^2=0-----------(equation number 3)

Let's solve equation number 1 first

(x+1)^2 =0

Taking square root in both sides -

x+1=0

x=(-1)------------(first root of the equation)

Let's solve the equation number 2

(x-2)^2 =0

x-2 =0

x =2-------------(second root of the equation)

Let's solve the equation number 3

(x-3)^2=0

x-3=0

x=3---------------(third root of the equation)

Thus the number of real roots of the given equation is 3

Answer 2

Answer:

0

Step-by-step explanation:

(x-1)² + (x - 2)² + (x - 3)² = 0

We know that a² + b² + 2 = 0 it is possible when

a = b = c = 0

x-1=0 and x-2=0 and x-3 = 0

x= 1 and x = 2 and x = 3 So there is no value of x for which

x-1= x-2 = x-3 = 0