Question

If the following term $a\left( b-c \right){{x}^{2}}+b\left( c-a \right)xy+c\left( a-b \right){{y}^{2}}$ is a perfect square, then the quantities $a,b,c$ are in
A) A.P
B) G.P
C) H.P
D) None of these

Answer 1

Hint: We start solving this problem by equating the discriminant of the given equation to zero as we were given that the quadratic expression is a perfect square. The formula we need to find the discriminant is ${{b}^{2}}-4ac$ for the equation $a{{x}^{2}}+bx+c$. Then we solve the discriminant and find the relation between a, b and c.

Complete step by step answer:
We were given that the equation $a\left( b-c \right){{x}^{2}}+b\left( c-a \right)xy+c\left( a-b \right){{y}^{2}}$ is a perfect square.
Any quadratic equation is said to be a perfect square if its roots are equal. So, the given equation has equal roots.
To find the relation between the coefficients of the equation, first let us go through the nature of the roots of a quadratic equation say $a{{x}^{2}}+bx+c$.
If the quadratic equation has two imaginary roots, then the discriminant is less than zero. ${{b}^{2}}-4ac<0$
If the quadratic equation has two real and equal roots, then the discriminant is equal to zero. ${{b}^{2}}-4ac=0$
If the quadratic equation has two real and distinct roots, then the discriminant is greater than zero.
${{b}^{2}}-4ac>0$
As we were given that the given expression is perfect square. So, it has two real and equal roots. So, its discriminant is equal to zero.
So, we use the above formula ${{b}^{2}}-4ac=0$.
Now, we apply this formula to the given expression, we get
\[\begin{align}
  & \Rightarrow {{\left( b\left( c-a \right) \right)}^{2}}-4\left( a\left( b-c \right) \right)\left( c\left( a-b \right) \right)=0 \\
 & \Rightarrow {{\left( b\left( c-a \right) \right)}^{2}}=4\left( a\left( b-c \right) \right)\left( c\left( a-b \right) \right) \\
 & \Rightarrow {{\left( bc-ab \right)}^{2}}=4ac\left( b-c \right)\left( a-b \right) \\
\end{align}\]
By simplifying this equation using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we get
$\begin{align}
  & \Rightarrow {{b}^{2}}{{c}^{2}}+{{a}^{2}}{{b}^{2}}-2a{{b}^{2}}c=4ac\left( ab-{{b}^{2}}-ac+bc \right) \\
 & \Rightarrow {{b}^{2}}{{c}^{2}}+{{a}^{2}}{{b}^{2}}-2a{{b}^{2}}c=4{{a}^{2}}bc-4a{{b}^{2}}c-4{{a}^{2}}{{c}^{2}}+4ab{{c}^{2}} \\
 & \Rightarrow {{b}^{2}}{{c}^{2}}+{{a}^{2}}{{b}^{2}}-2a{{b}^{2}}c-4{{a}^{2}}bc+4a{{b}^{2}}c+4{{a}^{2}}{{c}^{2}}-4ab{{c}^{2}}=0 \\
 & \Rightarrow {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+4{{a}^{2}}{{c}^{2}}+4a{{b}^{2}}c-4{{a}^{2}}bc-4ab{{c}^{2}}=0 \\
 & \Rightarrow {{\left( ab \right)}^{2}}+{{\left( bc \right)}^{2}}+{{\left( -2ac \right)}^{2}}+2\left( ab \right)\left( bc \right)+2\left( -2ac \right)\left( ab \right)+2\left( bc \right)\left( -2ac \right)=0...............\left( 1 \right) \\
\end{align}$
Now let us the formula,
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$
As the expression in equation (1) resembles the same, we can use the above formula and write the equation (1) as,
$\begin{align}
  & \Rightarrow {{\left( ab+bc-2ac \right)}^{2}}=0 \\
 & \Rightarrow \left( ab+bc-2ac \right)=0 \\
 & \Rightarrow ab+bc=2ac \\
 & \Rightarrow b\left( a+c \right)=2ac \\
 & \Rightarrow b=\dfrac{2ac}{a+c} \\
\end{align}$
We all know that if $a,b,c$ are H.P, then the relation between $a,b,c$ is
$b=\dfrac{2ac}{a+c}$
So, we can say that $a,b,c$ are in H.P.
So, if $a\left( b-c \right){{x}^{2}}+b\left( c-a \right)xy+c\left( a-b \right){{y}^{2}}$ is a perfect square, then the quantities $a,b,c$ are in H.P.

So, the correct answer is “Option C”.

Note: There is another method for solving this question. Let us consider the given equation.
$a\left( b-c \right){{x}^{2}}+b\left( c-a \right)xy+c\left( a-b \right){{y}^{2}}$
Let us divide the equation with ${{y}^{2}}$, then we get
$a\left( b-c \right){{\left( \dfrac{x}{y} \right)}^{2}}+b\left( c-a \right)\left( \dfrac{x}{y} \right)+c\left( a-b \right)$
Let us substitute $\dfrac{x}{y}=1$ in the above equation. Then the value of the equation is,
$\begin{align}
  & \Rightarrow a\left( b-c \right){{\left( 1 \right)}^{2}}+b\left( c-a \right)\left( 1 \right)+c\left( a-b \right) \\
 & \Rightarrow a\left( b-c \right)+b\left( c-a \right)+c\left( a-b \right) \\
 & \Rightarrow ab-ac+bc-ba+ac-bc \\
 & \Rightarrow 0 \\
\end{align}$
So, 1 is a root of the equation. As we were given that it is a perfect square it must have equal roots. So, our obtained equation has roots 1 and 1.
So, product of the roots is 1, as product of 1 and 1 is 1.
So, consider the product of the roots for the equation $a\left( b-c \right){{\left( \dfrac{x}{y} \right)}^{2}}+b\left( c-a \right)\left( \dfrac{x}{y} \right)+c\left( a-b \right)$ and let us equate it to 1.
$\begin{align}
  & \Rightarrow \dfrac{c\left( a-b \right)}{a\left( b-c \right)}=1 \\
 & \Rightarrow c\left( a-b \right)=a\left( b-c \right) \\
 & \Rightarrow ac-bc=ab-ac \\
 & \Rightarrow ab+bc=2ac \\
 & \Rightarrow b\left( a+c \right)=2ac \\
 & \Rightarrow b=\dfrac{2ac}{a+c} \\
\end{align}$
So, we get that a, b, c are in H.P.
Hence the answer is Option C.