Question

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3. 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.
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Answer 1

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

A = 3 B = 2( a + b + c )

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

Hence, the roots are equal only when

a=b=c