Question

A solution of the quadratic equation (a+b–2c)x2
+ (2a–b–c)x + (c+a–2b) = 0 is

Answer 1

Answer:

⇒ $x=\frac{-(a - 2b +c)}{a + b - 2c}$ (or) ⇒ $x= -1$

Step-by-step explanation:

Given :

The solution for quadratic equation :

(a + b – 2c)x² + (2a – b – c)x + (c + a – 2b) = 0.

Solution :

(a + b – 2c)x² + (2a – b – c)x + (a – 2b + c) = 0

By using Quadratic formula,

⇒ $x=\frac{-b±\sqrt{b^2 - 4ac} }{2a}$

Substituting a = (a + b - 2c) , b = (2a - b - c) , c = (a - 2b + c)

⇒ $x=\frac{-(2a - b - c)±\sqrt{(2a - b - c)^2 - 4(a + b - 2c) (a - 2b + c)} }{2(a + b - 2c)}$

⇒ $x=\frac{b + c - 2a±\sqrt{(4a^2 - 4ab - 4ac + b^2 + 2bc + c^2) - 4(a^2 - ab - ac-2b^2+5bc-2c^2)} }{2(a + b - 2c)}$

⇒ $x=\frac{b + c - 2a±\sqrt{(4a^2 - 4ab - 4ac + b^2 + 2bc + c^2) - (4a^2 - 4ab - 4ac-8b^2 +20bc - 8c^2)} }{2(a + b - 2c)}$

⇒ $x=\frac{b + c - 2a±\sqrt{9b^2 - 18bc + 9c^2} }{2(a + b - 2c)}$

⇒ $x=\frac{b + c - 2a±\sqrt{9(b^2 - 2bc + c^2)} }{2(a + b - 2c)}$

⇒ $x=\frac{b + c - 2a±\sqrt{[3(b - c)]^2} }{2(a + b - 2c)}$

⇒ $x=\frac{b + c - 2a+[3(b - c)] }{2(a + b - 2c)}$ (or) ⇒ $x=\frac{b + c - 2a-[3(b - c)] }{2(a + b - 2c)}$

⇒ $x=\frac{b + c - 2a+[3b - 3c] }{2(a + b - 2c)}$ (or) ⇒ $x=\frac{b + c - 2a-[3b -3c] }{2(a + b - 2c)}$

⇒ $x=\frac{b + c - 2a+3b - 3c }{2(a + b - 2c)}$ (or) ⇒ $x=\frac{b + c - 2a-3b+3c] }{2(a + b - 2c)}$

⇒ $x=\frac{-2a + 4b - 2c}{2(a + b - 2c)}$ (or) ⇒ $x=\frac{-2a -2b+4c}{2(a + b - 2c)}$

⇒ $x=\frac{-2(a - 2b +c)}{2(a + b - 2c)}$ (or) ⇒ $x=\frac{-2(a +b-2c)}{2(a + b - 2c)}$

⇒ $x=\frac{-(a - 2b +c)}{a + b - 2c}$ (or) ⇒ $x= -1$