.Prove that both the roots of the equation (x – a)(x – b)+(x – b)(x – c) + (x –
c)(x – a) = 0 are real but they are equal only when a = b = c.
Answers
Step-by-step explanation:
Solution :-
Given that
(x - a)(x - b)+(x - b)(x - c) + (x -c)(x-a) = 0
=> x²-(a+b)x+ab + x²-(b+c)x+bc+x²-
(c+a)x+ac = 0
=> 3x²-(a+b+b+c+c+a)x+ab+bc+ca = 0
=> 3x²-(2a+2b+2c)x+ab+bc+ca = 0
=> 3x²-2(a+b+c)x +ab+bc+ca = 0
We know that
The discriminant of ax²+bx+c = 0 is b²-4ac
If b²-4ac = 0 then the roots are real and equal.
=> [-2(a+b+c)]²-4(3)(ab+bc+ca) = 0
=>4(a²+b²+c²+2ab+2bc+2ca)-12(ab+bc+ca) = 0
=>4[a²+b²+c²+2ab+2bc+2ca-3(ab+bc+ca)]= 0
=>a²+b²+c²+2ab+2bc+2ca-3(ab+bc+ca)
= 0
=> a²+b²+c²+2ab+2bc+2ca-3ab-3bc-3ca
= 0
=> a²+b²+c²-ab-bc-ca = 0
=> a²+b²+c² = ab+bc+ca
On multiplying 2 both sides
=>2( a²+b²+c²) = 2(ab+bc+ca)
=> 2a²+2b²+2c² = 2ab+2bc+2ca
=> a²+a²+b²+b²+c²+c² = 2ab+2bc+2ca
=> a²+a²+b²+b²+c²+c² -2ab-2bc-2ca = 0
=> (a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²) = 0
We know that
(x-y)² = x²-2xy+y²
=> (a-b)²+(b-c)²+(c-a)² = 0
=> (a-b)² = 0 or (b-c)² = 0 or (c-a)² = 0
=> a-b = 0 or b-c = 0 or c-a = 0
=> a = b or b = c or c = a
=> a = b = c
The roots of the equation (x - a)(x - b)+(x - b)(x - c) + (x -c)(x-a) = 0 are real and equal only when a = b = c
Hence, Proved.
PROVIDED INFORMATION :-
- Prove that both the roots of the equation
- (x – a)(x – b)+(x – b)(x – c) + (x –c)(x – a) = 0
QUESTION :-
Prove that both the roots of the equation (x – a)(x – b)+(x – b)(x – c) + (x –c)(x – a) = 0 are real but they are equal only when a = b = c
GIVEN :-
- (x – a)(x – b)+(x – b)(x – c) + (x –c)(x – a) = 0
TO PROVE :-
- a = b = c
SOLUTION :-
The given equation may be written as
3x ^ 2 - 2(a + b + c) * x + (ab + bc + ac) = 0
Discriminant D = B² - 4AC
D = 4 * (a + b + c) ^ 2 - 4 * 3 * (ab + bc + ac)
D = 4(a ^ 2 + b ^ 2 + c ^ 2 + 2ab + 2bx + 2ac)
12(ab + bc + ac)
= 4(a²+ b² + 2 -ab -bc - ac)
= 2(2a²+ 2b² + 2² - 2ab - 2 bc - 2ac)
= 2[a²+ b²- 2ab + b² +c²- 2bc + c² a² - 2ac]
= 2[(a - b) ^ 2 + (b - c) ^ 2 + (c - a) ^ - 2] >= 0
= (a - b)² ≥ 0, (b-c) ² ≥ 0 and (c-a)² ≥ 0]
Hence, both roots of the equation are
real.
For equal root we must have D = 0
= (a - b)²+(b-c)² + (c-a)² = 0
a - b=0, b - c =0 , c - a =0
a = b, b = c , c = a
a = b = c