Question

If roots of the quadratic equation $\left( b-c \right){{x}^{2}}+\left( c-a \right)x+\left( a-b \right)=0$ are real and equal then prove that $2b=a+c$.

Answer 1

We solve this problem by first considering the concept of nature of the roots of a quadratic equation. Then we use the formula for the determinant of the quadratic equation $a{{x}^{2}}+bx+c$, ${{b}^{2}}-4ac$ and equate it to zero as the roots are equal and real. Then we use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ for simplifying the obtained equation. Then we simplify it using the formula, ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac={{\left( a+b+c \right)}^{2}}$ and solve it to prove the required relation.

Complete step-by-step solution
We are given that the roots of the quadratic equation $\left( b-c \right){{x}^{2}}+\left( c-a \right)x+\left( a-b \right)=0$ are real and equal.
First, let us go through the nature of the roots of any quadratic equation say $a{{x}^{2}}+bx+c$.
If the quadratic equation has two imaginary roots, then the discriminant is less than zero, that is,
${{b}^{2}}-4ac<0$
If the quadratic equation has two real and equal roots, then the discriminant is equal to zero, that is,
${{b}^{2}}-4ac=0$
If the quadratic equation has two real and distinct roots, then the discriminant is greater than zero, that is,
${{b}^{2}}-4ac>0$
As we are given that the given equation $\left( b-c \right){{x}^{2}}+\left( c-a \right)x+\left( a-b \right)=0$ has equal roots, so discriminant is equal to zero.
Now, we apply this formula to the given expression, we get
$\Rightarrow {{\left( c-a \right)}^{2}}-4\left( b-c \right)\left( a-b \right)=0$
Now let us consider the formula,
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
By simplifying the above equation using this formula, we get,

& \Rightarrow {{\left( c-a \right)}^{2}}-4\left( b-c \right)\left( a-b \right)=0 \\\ & \Rightarrow \left( {{c}^{2}}+{{a}^{2}}-2ac \right)-4\left( ab-{{b}^{2}}-ac+bc \right)=0 \\\ & \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac-4ab+4{{b}^{2}}+4ac-4bc=0 \\\ & \Rightarrow {{c}^{2}}+{{a}^{2}}+4{{b}^{2}}-4ab+2ac-4bc=0 \\\ & \Rightarrow {{a}^{2}}+4{{b}^{2}}+{{c}^{2}}-4ab+2ac-4bc=0 \\\ \end{aligned}$$ We can write the above equation as, $$\begin{aligned} & \Rightarrow {{a}^{2}}+{{\left( 2b \right)}^{2}}+{{c}^{2}}-2\left( a \right)\left( 2b \right)+2\left( a \right)\left( c \right)-2\left( 2b \right)\left( c \right)=0 \\\ & \Rightarrow {{a}^{2}}+{{\left( -2b \right)}^{2}}+{{c}^{2}}+2\left( a \right)\left( -2b \right)+2\left( a \right)\left( c \right)+2\left( -2b \right)\left( c \right)=0 \\\ \end{aligned}$$ Now let us consider the formula, $${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac={{\left( a+b+c \right)}^{2}}$$ Using this formula, we can write the above equation as, $$\begin{aligned} & \Rightarrow {{a}^{2}}+{{\left( -2b \right)}^{2}}+{{c}^{2}}+2\left( a \right)\left( -2b \right)+2\left( a \right)\left( c \right)+2\left( -2b \right)\left( c \right)=0 \\\ & \Rightarrow {{\left( a-2b+c \right)}^{2}}=0 \\\ & \Rightarrow a-2b+c=0 \\\ & \Rightarrow 2b=a+c \\\ \end{aligned}$$ So, we get that $2b=a+c$. Hence Proved. **Note:** We can also solve this question in another method. Let us substitute the value $x=1$ in the given equation. Then we get, $\begin{aligned} & \Rightarrow \left( b-c \right){{\left( 1 \right)}^{2}}+\left( c-a \right)\left( 1 \right)+\left( a-b \right)=0 \\\ & \Rightarrow \left( b-c \right)+\left( c-a \right)+\left( a-b \right)=0 \\\ & \Rightarrow 0=0 \\\ \end{aligned}$ So, we get that 1 is a root. As the equation has equal roots both the roots are 1. Now let us consider the product of the roots of the given equation, $\left( b-c \right){{x}^{2}}+\left( c-a \right)x+\left( a-b \right)=0$. $\begin{aligned} & \Rightarrow \dfrac{a-b}{b-c}=1\times 1=1 \\\ & \Rightarrow a-b=b-c \\\ & \Rightarrow 2b=a+c \\\ \end{aligned}$ Hence Proved.