If $a > b > c$, then equation $2 ( x - a ) ( x - b ) - ( x - c ) ^ { 2 } = 0$ has-
- exactly one root in $(b,a) $
- exactly one root in $(c,b)$
- both in $(c ,a)$
- one root in $( - \infty , b )$ and other in $( a , \infty )$
Let f(x) = 2(x-a)(x-b) - (x-c)^2. f(c) = 2(c-a)(c-b) = 2(negative)(negative) > 0. f(b) = -(b-c)^2 < 0. Since f(c) > 0 and f(b) < 0, there is a root in (c,b). f(a) = -(a-c)^2 < 0. Since f(b) < 0 and f(a) < 0, we check limits. As x -> infinity, f(x) -> x^2 > 0. Thus there is another root in (a, infinity).
Let f of x equal 2 times the quantity x minus a times the quantity x minus b minus the square of the quantity x minus c. At x equals b, f of b equals negative b minus c squared, which is less than zero because b is greater than c. At x equals c, f of c equals 2 times c minus a times c minus b, which is positive because both factors are negative. By the Intermediate Value Theorem, f of x must equal zero for exactly one root between c and b.