Question

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

(a) 2

(b) 1

(c) 0

(d) 3

Please Explain.

Answer 1

Note:

(1) b^2 - 4ac > 0  ----- Positive      ---- 2 (Number of real roots)

(2) b^2 - 4ac < 0  ---- 0                 ---- 1(Number of real roots)

(3) b^2 - 4ac > 0  ----- Negative  ------ 0(Number of real roots).

Now,

 Given Equation is (x - 1)^2 + (x - 2)^2 + (x - 3)^2 = 0

= > x^2 + 1 - 2x + x^2 + 4 - 4x + x^2 + 9 - 6x = 0

= > 3x^2 - 12x + 14 = 0 .

Given that the equation has real roots.

= > b^2 - 4ac 

= > (12)^2 - 4(3)(14)

= > 144 - 168

= > -24

The discriminant is -ve, Therefore it has Zero real roots.

Hope this helps!

Answer 2

Answer:

Option (c) is correct

it has no real roots

Step-by-step explanation:

Step - 1

(x-1)²+(x-2)²+(x-3)² add all this eq.

i.e

(x²-2x+1)+(x²-4x+4)+(x²-6x+9) =0

3x²-12x+14

Step -2

Find it's Discriminate (D)

D=b²-4ac

Here a=3 , b=(-12) , c=14

D= (-12)²-4(3)(14)

D= 144-168

D=(-24)

Step 3

Compare it's Discriminate

D<0

(It is a statement that if D<0 it has no real roots and all roots are imaginary )

hence , this eq. has no real roots