Question

6. If the roots of the Quadratic equations (a-b)x2 + (b-c) x +(c-a ) = 0 are equal. Prove that 2a = b + c

Answer 1

Answer:

if the roots of the equation are equal then,

${b}^{2} - 4ac = 0$

${(b - c)}^{2} - 4 \times (a - b) \times (c - a) = 0$

$( {b}^{2} - 2bc + {c}^{2} ) - (4 \times (ac - {a}^{2} - bc + ab)) = 0$

$( {b}^{2} - 2bc + {c}^{2} ) - (4 \times ((a - b)c - {a}^{2} + ab )) = 0$

$( {b}^{2} - 2bc + {c}^{2} ) - (4(a - b)c - 4 {a}^{2} + 4ab ) = 0$

${b}^{2} - 2bc + {c}^{2} - 4ac + 4bc + 4 {a}^{2} - 4ab = 0$

${b}^{2} + 2bc + {c}^{2} - 4ac + 4 {a}^{2} - 4ab = 0$

$4 {a}^{2} + {b}^{2} + {c}^{2} - 4ab + 2bc - 4ac = 0$

${(2a)}^{2} + {( - b)}^{2} + {( - c)}^{2} + 2 \times 2a \times ( - b) + 2 \times ( - b) \times ( - c) + 2 \times 2a \times ( - c) = 0$

${(2a - b - c)}^{2} = 0$

$(2a - b - c) = \sqrt{0} = 0$

$2a = b + c$

hence, proven.

Answer 2

plz refer to this attachment