Question

If the equation (b - c) {x}^{2} + (c - a)x + (a - b) = 0 के मूल समान है । तो सिद्ध किजिए कि 2b = a + c​

Answer 1

$\mathfrak{\large{\underline{\underline{Given :-}}}}$

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

$\mathfrak{\large{\underline{\underline{To prove:-}}}}$

2b = a + c

$\mathfrak{\large{\underline{\underline{proof:-}}}}$

Let a = (b - c) , b = ( c - a) , c = (a - b)

Finding,it's discriminant by using discriminant rule :-

★ $\boxed{\sf{D ={b}^{2} - 4ac }}$

put the value of a, b, c

$\implies$ $\bold{D = {(c - a)}^{2} - 4.(b - c).(a - b)}$

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

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

Using suitable identity :-

$\boxed{\sf{{(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + \:2 bc + \: 2ca}}$

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

$\implies$ $\bold{2b = a + c}$