Question

Let $P,Q,R$ be defined as

& P={{a}^{2}}b+a{{b}^{2}}-{{a}^{2}}c-a{{c}^{2}} \\\ & Q={{b}^{2}}c+b{{c}^{2}}-{{a}^{2}}b-a{{b}^{2}} \\\ & R={{c}^{2}}a+c{{a}^{2}}-{{c}^{2}}b-c{{b}^{2}} \\\ \end{aligned}$$ where $a,b,c$ are all $+ive$ and the equation $P{{x}^{2}}+Qx+R=0$ has equal roots then $a,b,c$ are in $$$$ A. A.P. $$$$ B. G.P.$$$$ C. H.P.$$$$ D. None of these$$$$

Answer 1

We take ${{a}^{2}},a$ common in $P$, ${{b}^{2}},b$ common in $Q$ and ${{c}^{2}},c$ common in $R$ to express $P,Q,R$ in cyclical form. We put the cyclical expression of $P,Q,R$ in $P{{x}^{2}}+Qx+R=0$ and using the given condition equate the discriminant to zero. We simplify the discriminant equation until we get a relation only in $a,b,c$. If the relation is $2b=a+c$ then $a,b,c$ are in arithmetic progression (AP), if the relation is ${{b}^{2}}=ac$ then $a,b,c$ are in geometric progression (GP) and if the relation is $\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}$ then $a,b,c$ are in harmonic progression (HP).$$$$

Complete step-by-step answer:
We know that the standard quadratic equation is given by $a{{x}^{2}}+bx+c=0$ where $a,b,c$ are real numbers with condition has equal roots when the discriminant $D$ satisfies the condition $D={{b}^{2}}-4ac=0$
We are given from the question three expressions of $P,Q,R$ respectively as

& P={{a}^{2}}b+a{{b}^{2}}-{{a}^{2}}c-a{{c}^{2}} \\\ & Q={{b}^{2}}c+b{{c}^{2}}-{{a}^{2}}b-a{{b}^{2}} \\\ & R={{c}^{2}}a+c{{a}^{2}}-{{c}^{2}}b-c{{b}^{2}} \\\ \end{aligned}$$ We try to express $P,Q,R$ in cyclical form by taking ${{a}^{2}},a$ common in $P$, ${{b}^{2}},b$ common in $Q$ and ${{c}^{2}},c$ common in $R$. We have, $$\begin{aligned} & P={{a}^{2}}\left( b-c \right)+a\left( {{b}^{2}}-{{c}^{2}} \right)={{a}^{2}}\left( b-c \right)+a\left( b-c \right)\left( b+c \right)=a\left( b-c \right)\left( a+b+c \right) \\\ & Q={{b}^{2}}\left( c-a \right)+b\left( {{c}^{2}}-{{a}^{2}} \right)={{b}^{2}}\left( c-a \right)+b\left( c-a \right)\left( c+a \right)=b\left( c-a \right)\left( a+b+c \right) \\\ & R={{c}^{2}}\left( a-b \right)+c\left( {{a}^{2}}-{{b}^{2}} \right)={{c}^{2}}\left( a-b \right)+c\left( a-b \right)\left( a+b \right)=c\left( a-b \right)\left( a+b+c \right) \\\ \end{aligned}$$ We are further given in the question that $a,b,c$ are all positive real numbers and the equation $P{{x}^{2}}+Qx+R=0$ has equal roots. We can put the value of $P,Q,R$ in the given quadratic equation $P{{x}^{2}}+Qx+R=0$ and have, $$\Rightarrow a\left( b-c \right)\left( a+b+c \right){{x}^{2}}+b\left( c-a \right)\left( a+b+c \right)x+c\left( a-b \right)\left( a+b+c \right)=0$$ Since $a,b,c$ are positive real numbers $a+b+c$ cannot be zero and we can divide $a+b+c$ in both side of the above equation to have, $$\Rightarrow a\left( b-c \right){{x}^{2}}+b\left( c-a \right)x+c\left( a-b \right)=0$$ The discriminant of the above quadratic equation is zero since the quadratic equation has equal roots. So we have $$\begin{aligned} & {{\left( b\left( c-a \right) \right)}^{2}}-4\times a\left( b-c \right)\times c\times \left( a-b \right)=0 \\\ & \Rightarrow {{b}^{2}}{{\left( c-a \right)}^{2}}-4ac\left( ab-{{b}^{2}}-ca+bc \right)=0 \\\ & \Rightarrow {{b}^{2}}{{c}^{2}}+{{b}^{2}}{{a}^{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}}+2a{{b}^{2}}c-4ab{{c}^{2}}-4{{a}^{2}}bc=0 \\\ & \Rightarrow {{\left( ab \right)}^{2}}+{{\left( bc \right)}^{2}}+{{\left( -2ac \right)}^{2}}+2\times ab\times bc+2\times bc\times \left( -2ac \right)+2\times ab\times \left( -2ac \right)=0 \\\ \end{aligned}$$ We use the algebraic identity ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$ for $x=ab,y=bc,z=-2ac$ in the above step to have $$\Rightarrow {{\left( ab+bc-2ac \right)}^{2}}=0$$ We take square root of both side of the above step to have, $$\begin{aligned} & \Rightarrow ab+bc-2ac=0 \\\ & \Rightarrow ab+bc=2ac \\\ \end{aligned}$$ Since $a,b,c$ are positive real numbers $abc$ cannot be zero and we can divide $abc$ in both side of the above equation to have, $$\Rightarrow \dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}$$ The above relation implies that $a,b,c$ are in HP. So the correct option C$$$$ **So, the correct answer is “Option C”.** **Note:** We can alternatively solve using the product of roots of $P{{x}^{2}}+Qx+R=0$ and equating it to a constant. The key in this problem is to express the discriminant into a square of $\left( ab+bc-2ac \right)$. Harmonic progression (HP) is a mathematical sequence formed from the reciprocals of the terms in an AP sequence.