Question

If the roots of the equation `(b-c)x^2+ (c-a)x+ (a-b) = 0` are equal, then how do you prove that `2b = a + c`?

Answer 1

See below.

Explanation:

Solving `(b-c)x^2+ (c-a)x+ (a-b) = 0` we have

`x = (-(c-a)pm sqrt((c-a)^2-4(b-c)(a-b)))/(2(b-c))`

the condition for equal roots is

`(c-a)^2-4(b-c)(a-b)=0` or

`(a-2b+c)^2=0` and then

`2b = a+c`

Answer 2

Because the roots of the equation are equal we know that:

`(x - r)(x-r) = 0`

where `r` is the value of the repeated root.

The equation can be written in the form:

`x^2 - 2rx+r^2 = 0" [1]"`

Given:

`(b-c)x^2+ (c-a)x+ (a-b) = 0`

Divide both sides by `(b-c)`:

`x^2+(c-a)/(b-c)x+(a-b)/(b-c) = 0" [2]"`

Matching the coefficients of the terms of equation [1] with the coefficients of the terms of equation [2], we obtain two non-trivial equations:

`-2r = (c-a)/(b-c)" [3]"`

`r^2 = (a-b)/(b-c)" [4]"`

Solve equation [3] for r:

`r = -1/2(c-a)/(b-c)" [3.1]"`

Substitute the right side of equation [3.1] into the left side of equation [4]:

`(-1/2(c-a)/(b-c))^2 = (a-b)/(b-c)`

`1/4(c-a)^2/(b-c)^2 = (a-b)/(b-c)`

`(c-a)^2= 4(a-b)(b-c)`

Perform the multiplication:

`c^2-2ac+a^2=4(ab-ac-b^2+bc)`

`c^2-2ac+a^2=4ab-4ac-4b^2+4bc`

Move everything to the left:

`c^2+2ac+a^2-4ab+4b^2-4bc= 0`

Factor by grouping:

`c(c+a-2b) +ac+a^2 +4b^2-2bc = 0`

Add 0 in the form of `-2ba+2ba`

`c(c+a-2b) +ac+a^2-2ba +4b^2-2bc + 2ba = 0`

`c(c+a-2b) +a(c+a-2b) +4b^2-2bc + 2ba = 0`

`c(c+a-2b) +a(c+a-2b) -2b(-2b^2+c + a) = 0`

The left side is a perfect square:

`(c+a-2b)^2 = 0`

Use the square root on both sides:

`c+a-2b = 0`

`2b = c+a` Q.E.D.