Math, asked by adharshm5740, 1 year ago

(a+b)x² +(a+2b+c)x+(b+c)=0

Answers

Answered by emmanuel2199
0

Step-by-step explanation:

(a+b-2c)X^2 + (a+b-2c)X +(a+c-2b) X + (a+c-2b) = 0

( a+b-2c)X { X+1} + (a+c-2b) {X+1} =0

(X+1) { (a+b-2c)X + ( a+c-2b)} =0

X = -1 or

(a+b-2c)X = -(a+b-2c)

=> X = -(a+c-2b)/(a+b-2c)

Answered by codiepienagoya
1

The value of X is \frac{-(b+c)}{(a+b)}  \   or \  -1\\

Step-by-step explanation:

(a+b)x^2 +(a+2b+c)x+(b+c)=0\\

ax^2+bx+c=0\\ comparing the value.

a=(a+b)\\

b=(a+2b+c)\\

c=(b+c)\\

formula: \frac{-b\pm \sqrt{b^2-4\cdot a \cdot c}}{2\cdot a}\\

x= \frac{-(a+2b+c)\pm \sqrt{(a+2b+c)^2-4\cdot (a+b) \cdot (b+c)}}{2\cdot(a+b)}\\

x= \frac{-(a+2b+c)\pm \sqrt{(a^2+(2b)^2+c^2+2\cdot a\cdot 2b+ 2\cdot 2b\cdot c+2 \cdot c\cdot a-4(ab+ac+b^2+bc)}}{2\cdot(a+b)}\\

x= \frac{-(a+2b+c)\pm \sqrt{(a^2+4b^2+c^2+4\cdot a\cdot b+ 4\cdot b\cdot c+2 \cdot c\cdot a-4ab-4ac-4b^2-4bc)}}{2a+2b}\\

x= \frac{-(a+2b+c)\pm \sqrt{(a^2+c^2 -2\cdot c\cdot a)}}{2a+2b}\\

x= \frac{-(a+2b+c)\pm \sqrt{(a^2+c^2 -2ca)}}{2a+2b}\\

x= \frac{-(a+2b+c)\pm \sqrt{(a-c)^2}}{2a+2b}\\

x= \frac{-(a+2b+c)\pm (a-c)}{2a+2b}\\

x= \frac{-a-2b-c\pm (a-c)}{2a+2b}\\

x= \frac{-a-2b-c+a-c}{2a+2b}\ or \  x= \frac{-a-2b-c-a+c}{2a+2b}\\

x= \frac{-2b-2c}{2a+2b} \ or \ x= \frac{-a-2b-a}{2a+2b}\\

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

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

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

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