Fsc Part 1 Mathematics (Complete Solution)

# Q19 Find roots of the following equations by using quadratic formula

(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0

Solution

x2 - bx - ax + ab + x2 - cx - bx + bc + x2 - ax - cx + ca = 0
3x2 - 2bx - 2ax - 2cx + ab + bc + ca = 0
3x2 - 2(a + b + c)x + (ab + bc + ca) = 0

Comparing the given equation with ax2 + bx + c = 0, we get
a = 3,
b = -2(a + b + c),
c = (ab + bc + ca)

By Quardratic formula, we have
x =
-b ±
 √ b2 - 4ac
2a
Putting the values of a, b, and c in the formula

x =
-[-2(a + b + c)] ±
 √ [-2(a + b + c)]2 - 4(3)(ab + bc + ca)
2(3)
x =
2(a + b + c) ±
 √ 4(a + b + c)2 - 4(3)(ab + bc + ca)
6
x =
2(a + b + c) ±
 √ 4[ (a + b + c)2 - (3)(ab + bc + ca) ]
6
x =
2(a + b + c) ±
 √ (2)2[ (a + b + c)2 - (3)(ab + bc + ca) ]
6
x =
2(a + b + c) ± 2
 √ (a + b + c)2 - (3)(ab + bc + ca)
6
x =
2(a + b + c) ± 2
 √ a2 + b2 + c2 + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca
6
x =
2(a + b + c) ± 2
 √ a2 + b2 + c2 - ab - bc - ca
6
x =
(a + b + c) ±
 √ a2 + b2 + c2 - ab - bc - ca
3
Either
x =
(a + b + c) -
 √ a2 + b2 + c2 - ab - bc - ca
3
OR
x =
(a + b + c) +
 √ a2 + b2 + c2 - ab - bc - ca
3

Hence Solution set =
{
(a + b + c) -
 √ a2 + b2 + c2 - ab - bc - ca
3
,
(a + b + c) +
 √ a2 + b2 + c2 - ab - bc - ca
3
}