Fsc Part 1 Mathematics (Complete Solution)

# Q20 Find roots of the following equations by using quadratic formula

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

Solution

Comparing the given equation with ax2 + bx + c = 0,
the Coefficient of x2 = (a + b)
the Coefficient of x = (a + 2b + c) and
the constant term = b + c

Putting the values in the Quardratic formula

x =
-(a + 2b + c) ±
 √ (a + 2b + c)2 - 4(a + b)(b + c)
2(a + b)
x =
-(a + 2b + c) ±
 √ a2 + 4b2 + c2 + 2(a)(2b) + 2(2b)(c) + 2ca - 4(ab + ca + b2 + bc)
2(a + b)
x =
-(a + 2b + c) ±
 √ a2 + 4b2 + c2 + 4ab + 4bc + 2ca - 4ab - 4ca - 4b2 - 4bc
2(a + b)
x =
-(a + 2b + c) ±
 √ a2 + c2 - 2ca
2(a + b)
x =
-(a + 2b + c) ±
 √ (a - c)2
2(a + b)
x =
 -(a + 2b + c) ± (a - c)
2(a + b)
Either
x =
 -(a + 2b + c) - (a - c)
2(a + b)
OR
x =
 -(a + 2b + c) + (a - c)
2(a + b)
x =
 -a - 2b - c - a + c 2(a + b)
x =
 -a - 2b - c + a - c 2(a + b)
x =
 -2a - 2b 2(a + b)
x =
 -2b - 2c 2(a + b)
x =
 -2(a + b) 2(a + b)
x =
 -2(b + c) 2(a + b)
 x = -1
x =
 -(b + c) (a + b)

Hence Solution set =
{
-1 , -
 (b + c) (a + b)
}