Question

If the roots of equation (b -c) x² + (c-a) X
+ (a - b) = 0 are equal prove that 2b = a + c.​

Answer 1

Answer:

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Answer 2

$\underline{ \underline{\huge{ \bf{ \red{question}}}}}$

If the roots of eqn (b-c) x² + (c - a) x + ( a- b) = 0 are equal , prove that 2 b = a + c .

$\underline{ \underline{ \huge{ \bf{ \red{knowledge \: required}}}}}$

In any quadratic eqn of the form

ax² + bx + c = 0

Discriminant represented by 'D' is given by

$\boxed{ \tt{discriminant = {b}^{2} - 4ac }}$

● If D < 0 then the quadratic eqn will have no real roots.

● If D = 0 then the quadratic eqn will have two real and equal roots.

● If D > 0 then quadratic eqn will have two distinct real roots.

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Know the algebraic identity

$\tt{ {(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2xz}$

$\underline{ \underline{ \huge{ \bf{ \red{solution}}}}}$

Given Quadratic eqn is

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

As given That roots of this eqn are equal

Therefore,

Discriminant = 0

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

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

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

Solving and

Rearranging the terms

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

Using algebraic identity

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

$\implies \sf{a - 2b + c = 0}$

$\implies \sf{ \purple{a + c = 2b}}$

Proved.