Question

If the quadratic equation is (x-a) (x-b) + (x-b) (x-6) + (x-c) (x-a)=0
equal then show that a = b = c

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

If the quadratic equation is (x-a) (x-b) + (x-b) (x-c) + (x-c) (x-a)=0equal then show that a = b = c

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\large\underline\purple{\bold{Solution }}$

$\sf \:  ⟼(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$

$\sf \:  {x}^{2} - ax - bx + ab + {x}^{2} - bx - cx + bc + {x}^{2} - cx - ax + ac = 0$

$\sf \:  {3x}^{2} - \: 2(a + b + c) + (ab + bc + ca) = 0$

☆ Its a quadratic equation, it implies

$\begin{gathered}\begin{gathered}\bf where= \begin{cases} &\sf{A = 3} \\ &\sf{B = - 2(a + b - c)}\\ &\sf{C = ab + bc + ca} \end{cases}\end{gathered}\end{gathered}$

☆Since, quadratic equation has equal roots.

☆ Therefore, Discriminant = 0

$\bf\implies \: {B}^{2} - 4AC = 0$

$\sf \:  {\bigg( - 2(a + b + c)\bigg)}^{2} - 4 \times 3 \times (ab + bc + ca) = 0$

$\bf\implies \: {\bigg( a + b + c\bigg)}^{2} - 3(ab + bc + ca) = 0$

$\sf \:  ⟼ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca = 0$

$\sf \:  ⟼ {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca = 0$

☆ Multiply by 2 on both sides, we get

$\sf \:  ⟼2 {a}^{2} + 2{b}^{2} + 2{c}^{2} - 2ab -2 bc -2 ca = 0$

$\sf \:  ⟼ {a}^{2} + {a}^{2} + {b}^{2} + {b}^{2} + {c}^{2} + {c}^{2} - 2ab - 2bc - 2ca = 0$

$\sf \:  ⟼( {a}^{2} + {b}^{2} - 2ab) + ( {b}^{2} + {c}^{2} - 2bc) + ( {c}^{2} + {a}^{2} - 2ac) = 0$

$\sf \:  ⟼ {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} = 0$

☆ Since, sum of squares = 0

$\bf\implies \:a - b = 0 \: and \: b - c = 0 \: and \: c - a = 0$

$\bf\implies \:a = b \: and \: b = c \: and \: c = a$

$\bf\implies \:a = b = c$

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$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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