Question

6. Show that the roots of the equation

Answer 1

(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0
⇒x² - (a + b)x + ab + x² - (b + c)x + bc + x² - (c + a)x + ca = 0
⇒ 3x² - 2(a + b + c)x + (ab + bc + ca) = 0
Now, roots will be real only when D ≥ 0
So,D = {2(a + b + c)}² - 4.3(ab + bc + ca)
= 4(a + b + c)² - 12(ab + bc + ca)
= a² + b² + c² -ab - bc - ca
= 1/2[2a² + 2b² + 2c² - 2ab - 2bc - 2ca ]
= 1/2[ (a - b)² + (b - c)² + (c - a)² ]
Hence, it is clear that D ≥ 0 so, roots of given equation are always real numbers
And they cannot be equal unless , a = b , b = c and c = a ;Means, a = b = c

Answer 2

Solution:

Given ,

(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0

=> x²-(a+b)x+ab+x²-(b+c)x+bc

+x²-(c+a)x+ac=0

=>3x² -(a+b+b+c+c+a)x+ab+bc+ca = 0

=> 3x²-2(a+b+c)x+ ( ab+bc+ca) = 0

Compare above equation with

Ax² + Bx + C = 0 we get

A = 3 , B = 2(a + b + c) , C =ab+bc+ca ,

Discreminant = B² - 4AC

= [2(a+b+c)]² - 4×3×(ab+bc+ca)

= 4(a+b+c)² - 12(ab+bc+ca)

= 4[(a+b+c)² - 3(ab+bc+ca )]

=4[a²+b²+c²+2(ab+bc+ca)-3(ab+bc+ca)]
=4[a²+b²+c²-(ab+bc+ca ) ]

= 4×{1/2[(a-b)²+(b-c)²+(c-a)²]

= 2[(a-b)²+(b-c)²+(c-a)²]

≥ 0

Therefore,

roots are always real .

unless a = b = c .

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