Question

If the roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c )(x-a)=0 are real and equal then show that a=b=c

Answer 1

SOLUTION

The given quadratic equation is

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

$= > {x}^{2} - ax - bx + ab + {x}^{2} - bx - cx + bc + {x}^{2} - cx - ax + ca = 0 \\ = > {3x}^{2} - 2(a + b + c)x + (ab + bc + ca) = 0 \\ \\ discriminant = ( - 2(a + b + c)) {}^{2} - 4 \times 3 \times (ab + bc + ca) \\ = > 4(a + b + c) {}^{2} - 12(ab + bc + ca) \\ = > 4( {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca) \\ = > 4( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ = > 2(2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} - 2ab - 2bc - 2ca) \\ = > 2(( {a}^{2} - 2ab + {b}^{2} ) + ( {b}^{2} - 2bc + {c}^{2} ) + ( {c}^{2} - 2ca + {a}^{2} )) \\ = > 2((a - b) {}^{2} + ( {b - c)}^{2} + ( {c - a)}^{2} ) \\ = > since \: (a - b) {}^{2} + (b - c) {}^{2} + (c - a) {}^{2} is \: always \: positive.$

D> 0. so, the roots the given quadratic equation are equal.

Where a= b= c, D = 2× 0 = 0

Thus, the roots of the given quadratic equation are equal.

hope it helps ☺️⬆️