Question

Prove that both the roots of the equation (x - a)(x - b) + (x -
b)(x-c)+(x-c)(x-a)=0
are real but they are equal only when a = b =c​

Answer 1

Answer:

you say that a=b=c

so a=c

and a=b

if we place of c and b change by'a' so equation is prove.

Answer 2

Step-by-step explanation:

We have,

°•° ( x - a ) ( x - b ) + ( x - b ) ( x - c ) + ( x - c ) ( x - a ) = 0 .

==> x² - bx - ax + ab + x² - cx - bx + bc + x² - ax - cx + ac = 0 .

==> 3x² - 2bx - 2ax - 2cx + ab + bc + ca = 0 .

==> 3x² - 2x( a + b + c ) + ( ab + bc + ca ) = 0 .

When equation is compared with Ax² + Bx + C = 0 .

Then , A = 3 .

B = 2( a + b + c ) .

And, C = ( ab + bc + ca ) .

•°• Discriminant ( D ) = 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 + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca ) .

= 4( a² + b² + c² - ab - bc - ca ) .

= 2( 2a² + 2b² + 2c² - 2ab - 2bc - 2ca ) .

= 2[ ( a - b )² + ( b - c )² + ( c - a )² ] ≥ 0 .

[ °•° ( a - b )² ≥ 0, ( b - c )² ≥ 0 and ( c - a )² ≥ 0 ] .

This shows that both the roots of the given equation are real .

For equal roots, we must have : D = 0 .

Now, D = 0 .

==> ( a - b )² + ( b - c )² + ( c - a )² = 0 .

==> ( a - b ) = 0, ( b - c ) = 0 and ( c - a ) = 0 .

$\huge \boxed{ \green{ \sf \implies a = b = c . }}$

......

✔✔ Hence, the roots are equal only when a = b = c ..✅✅