If the roots of the equation (b-c)x^2+(c-a)x+(a-b)=0 are equal, prove that 2b=a+c.
The discriminant must be zero, so
(c-a)^2-4(b-c)(a-b)=0
c^2+a^2−2ac−4(ab−b^2−ac+bc)=0
c^2+a^2−2ac−4ab+4b^2+4ac−4bc=0
c^2+a^2+2ac+4b^2−4ab−4bc=0
(c+a)^2−4b(a+c)+4b^2=0
(c+a)^2−2⋅(c+a)⋅(2b)+(2b)^2=0
[(c+a)−(2b)]^2=0
c+a−2b=0
2b = a+c
To prove that 2b = a + c when the roots of the equation (b-c)x^2 + (c-a)x + (a-b) = 0 are equal, we need to use the property of quadratic equations with equal roots.
Let's call the common root of the equation as 'r'. The quadratic equation can then be written as:
(b-c)x^2 + (c-a)x + (a-b) = (x-r)(x-r) = (x-r)^2
Expanding the right side using the formula (a-b)^2 = a^2 - 2ab + b^2, we get:
(x-r)^2 = x^2 - 2rx + r^2
Comparing the coefficients of x^2, x, and the constant term between the original equation and the expanded form, we have:
Coefficient of x^2: b - c = 1
Coefficient of x: c - a = -2r
Constant term: a - b = r^2
Since the roots are equal, we know that c - a = -2r = 0. Therefore, r = 0, and consequently, a = c.
Substituting a = c into the constant term equation, we have:
a - b = 0^2
a - b = 0
a = b
Finally, substituting a = b into the equation 2b = a + c, we get:
2b = b + c
2b = a + c
Hence, we have proved that 2b = a + c when the roots of the given equation are equal.