QUADRATIC EQUATIONS
If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal,
then prove that 2b = a + c.
Sol.
(b – c)x2 + (c – a)x + (a – b) = 0
On comparing with Ax2 + Bx + C = 0, we get
A = (b – c),
B = (c – a),
C = (a – b)
Since, given equation has equal roots
∴ B2 – 4AC = 0
∴ (c – a)2
–
4
(b – c)
(a – b)
= 0
∴ c2
– 2ac
+ a2
– 4
(ab
– b2
– ac
+ bc)
= 0
∴ c2
– 2ac
+ a2
– 4ab
+ 4b2
+ 4ac
– 4bc
= 0
∴
(a + c)2
– 4b
(a + c)
+ (2b)2
= 0
∴ [(a + c)
–
(2b)
]2
= 0
Taking square roots on both sides
∴ (a + c) – (2b)
= 0
∴ a + c = 2b
Hence proved
What we need to prove?
What is given?
Means
B2 – 4AC = 0
Lets substitute the values
Since the coefficients in the equation are small alphabet a, b, c
We will write the standard form using Capital Alphabets A, B, C
Use
(x – y)2 = x2 – 2xy + y2
∴ c2
+ 2ac
+ a2
– 4ab
+ 4b2
– 4bc
= 0
We know
a2 + 2ac + c2 = (a + c)2
Rearranging the terms
∴ a2
+ 2ac
+ c2
– 4ab
– 4bc
+ 4b2
= 0
4b2 = (2b)2
x2
y2
x2 – 2xy + y2
= (x – y)2
Take – 4b common
– 2xy
– 2(a + c)2b
x = (a + c)
y = 2b