Question

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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

▶ 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 .

$\huge \pink{ \mid{ \underline{ \overline{ \sf Solution :- }} \mid}}$

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 . .

Answer 2

Solution:

We know that:

Root of quadratic equation,

(x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c) = 0 are equal.

That means:

$\implies$ $\huge{\boxed{\sf{D = b^{2} - 4ac}}}$

Now:

$\implies$ x² - (a + b)x + ab + x² - (b + c)x + bc + x² - (c + a)x + ca

$\implies$ 3x² - 2(a + b + c)x + (ab + bc + ca)

And:

D = {2(a + b + c)}² - 4(ab + bc + ca).3 = 0

$\implies$ 4{a² + b² + c² + 2(ab + bc + ca)} -12(ab + bc + ca) = 0

$\implies$ a² + b² + c² - ab - bc - ca = 0

$\implies$ 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

$\implies$ (a - b)² + (b - c)² + (c - a)² = 0

This is possible only when,

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

Hence proved,

If roots of given equation are equal then:

$\implies$ $\boxed{\sf{a = b = c}}$