Question

If roots of equation $\left(b-c\right)x^2+\left(c-a\right)x+\left(a-b\right)=0$ are equal then show that $2b=a+c$.

Answer 1

Step 1:

Compare the given equation with the standard quadratic equation.

The standard quadratic equation is $A{x}^2+Bx+C$.

Comparing both the equations, we get,

$\begin{array}{rcl}A& =& (b-c)\\ B& =& (c-a)\\ C& =& (a-b)\end{array}$

Step 2:

Prove the required relation:

If the roots of a quadratic equation are equal, the discriminant of the quadratic equation is zero.

The discriminant is calculated as,

$\begin{array}{rcl}D& =& B^2-4AC\\ & =& {\left(c-a\right)}^2-\left(4\times \left(b-c\right)\times \left(a-b\right)\right)\\ & =& {\left(c-a\right)}^2-4\left(b-c\right)\left(a-b\right)\end{array}$

Equate the discriminant to zero.

$$\begin{gathered} {\left(c-a\right)}^2-4(b-c)(a-b)=0 \\ \Rightarrow c^2+a^2-2ac-4(ab-b^2-ac+bc)=0 \\ \Rightarrow c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0 \\ \Rightarrow c^2+a^2+2ac+4b^2-4ab-4bc=0 \\ \Rightarrow {(c+a)}^2+4b^2-4b(a+c)=0 \\ \Rightarrow {(c+a)}^2+{\left(2b\right)}^2-2(c+a)\left(2b\right)=0 \\ \Rightarrow {\left[(c+a)-\left(2b\right)\right]}^2=0 \\ \Rightarrow c+a-2b=0 \\ \therefore 2b=c+a \end{gathered}$$

Hence, it is proved that if the roots of the equation $\left(b-c\right)x^2+\left(c-a\right)x+\left(a-b\right)=0$ are equal, then, $2b=a+c$.