Question

Let \[a,b,c,d\] be real numbers in G.P. If $u,v,w$ satisfy the system of equations$u+2v+3w=6$, $4u+5v+6w=12$, $6u+9v=4$. Then the roots of the equation $\left( \dfrac{1}{u}+\dfrac{1}{v}+\dfrac{1}{w} \right){{x}^{2}}+\left[ {{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(d-b)}^{2}} \right]x+u+v+w=0$ and $20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0$ are
(a) Equal
(b) Imaginary
(c) Reciprocals
(d) None of these

Answer 1

Hint:First, solve the given system of equations $u,v,w$ to obtain the values of the variables$u,v,w$ . Substitute these values in the given quadratic equation, whose roots are to be compared. If a, b, c, d are in G.P, then ${{b}^{2}}=ac$and ${{c}^{2}}=bd$ , $\dfrac{b}{a}=\dfrac{d}{c}$ . A real number can be called a root of a quadratic equation, if it satisfies the equation when substituted.

Complete step by step answer:

We are given a system of three equations which consist of three different variables $u,v$ and$w$.
$u+2v+3w=6........................(1)$
$4u+5v+6w=12.....................(2)$
$6u+9v=4...........................(3)$
We are also given two equations,
$\left( \dfrac{1}{u}+\dfrac{1}{v}+\dfrac{1}{w} \right){{x}^{2}}+\left[ {{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(d-b)}^{2}} \right]x+u+v+w=0................(4)$
$20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0.....................(5)$
We first have to solve for $u,v$and$w$.
Subtracting equation (1) from equation (2), we get,
$u+v+w=2....................(6)$
Further, we multiply equation (6) by 3 and then subtract equation (1) from it.
\[\begin{align}
  & \,\,\,\,3u+3v+3w=6 \\
 & -(\underline{1u+2v+3w=6}) \\
 & \,\,\,\,\underline{\underline{2u+1v+0w=0}} \\
\end{align}\]
We get,
$2u+v=0$
Or $u=\dfrac{-v}{2}$
Substituting $u=\dfrac{-v}{2}$ in equation (3), we get,
$6\left( \dfrac{-v}{2} \right)+9v=4$
$-3v+9v=4$
$6v=4$
$\therefore v=\dfrac{4}{6}=\dfrac{2}{3}$
$\therefore ~~u=\dfrac{-v}{2}=\dfrac{\left( -\dfrac{2}{3} \right)}{2}=\dfrac{-2}{6}=\dfrac{-1}{3}$
Substituting values for $u$ and $v$ in$u+v+w=2$, we get,
$w=2-u-v$
$=2-\left( \dfrac{-1}{3} \right)-\left( \dfrac{2}{3} \right)$
$=\dfrac{6}{3}+\dfrac{1}{3}-\dfrac{2}{3}$
$=\dfrac{5}{3}$
$\therefore w=\dfrac{5}{3}$
Hence, the values of $u,v$and$w$are $\dfrac{-1}{3}$, $\dfrac{2}{3}$ and $\dfrac{5}{3}$ respectively.
Substituting the values of $u,v$ and$w$ in equation (4), we get,
$\left( \dfrac{-3}{1}+\dfrac{3}{2}+\dfrac{3}{5} \right){{x}^{2}}+\left[ {{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(d-b)}^{2}} \right]x+\left( \dfrac{-1}{3}+\dfrac{2}{3}+\dfrac{5}{3} \right)=0$
$\left( \dfrac{-30}{10}+\dfrac{15}{10}+\dfrac{6}{10} \right){{x}^{2}}+\left[ {{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(d-b)}^{2}} \right]x+\left( \dfrac{6}{3} \right)=0$
$\left( \dfrac{-9}{10} \right){{x}^{2}}+\left[ {{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(d-b)}^{2}} \right]x+2=0$
Expanding ${{(b-c)}^{2}}$, ${{(c-a)}^{2}}$and ${{(d-b)}^{2}}$using the algebraic identity ${{(A-B)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$,
 we get,
$\left( \dfrac{-9}{10} \right){{x}^{2}}+\left[ {{b}^{2}}-2bc+{{c}^{2}}+{{c}^{2}}-2ca+{{a}^{2}}+{{d}^{2}}-2db+{{b}^{2}} \right]x+2=0$
$\left( \dfrac{-9}{10} \right){{x}^{2}}+\left[ 2{{b}^{2}}+2{{c}^{2}}+{{a}^{2}}+{{d}^{2}}-2bc-2ca-2db) \right]x+2=0...........(7)$
We are given that \[a,b,c,d\] be real numbers in G.P.
$\therefore {{b}^{2}}=ac$and ${{c}^{2}}=bd$
Also, $\dfrac{b}{a}=\dfrac{d}{c}$(= common ratio)\[^{2}\]
$\therefore bc=ad$
Substituting$ac={{b}^{2}}$, \[bd={{c}^{2}}\] and $bc=ad$ in equation (7), we get,
$\left( \dfrac{-9}{10} \right){{x}^{2}}+\left[ 2{{b}^{2}}+2{{c}^{2}}+{{a}^{2}}+{{d}^{2}}-2ad-2{{b}^{2}}-2{{c}^{2}}) \right]x+2=0$
$\left( \dfrac{-9}{10} \right){{x}^{2}}+\left[ {{a}^{2}}-2ad+{{d}^{2}} \right]x+2=0$
Now using the algebraic identity \[{{A}^{2}}-2AB+{{B}^{2}}={{(A-B)}^{2}}\],
$\left( \dfrac{-9}{10} \right){{x}^{2}}+{{(a-d)}^{2}}x+2=0$
To make the denominators same multiplying throughout by -10, we get,
$9{{x}^{2}}-10{{(a-d)}^{2}}x-20=0.....................(8)$
We have to find the relationship between roots of the equations $9{{x}^{2}}-10{{(a-d)}^{2}}x-20=0$ and $20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0$.
Let $\alpha $, $\beta $ be the roots of the equation $9{{x}^{2}}-10{{(a-d)}^{2}}x-20=0$.
Therefore,
$9{{\alpha }^{2}}-10{{(a-d)}^{2}}\alpha -20=0.....................(9)$
$9{{\beta }^{2}}-10{{(a-d)}^{2}}\beta -20=0.....................(10)$
Dividing equation (9) by ${{\alpha }^{2}}$, we get,
\[9-\dfrac{10{{(a-d)}^{2}}}{\alpha }-\dfrac{20}{{{\alpha }^{2}}}=0\]
\[9-10{{(a-d)}^{2}}\left( \dfrac{1}{\alpha } \right)-20\left( \dfrac{1}{{{\alpha }^{2}}} \right)=0\]
Multiplying by -1,we get,
\[20\left( \dfrac{1}{{{\alpha }^{2}}} \right)+10{{(a-d)}^{2}}\left( \dfrac{1}{\alpha } \right)-9=0\]
Comparing the above equation with $20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0$we can clearly see that $\dfrac{1}{\alpha }$ is a root of $20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0$.
Similarly,
Dividing equation (10) by ${{\beta }^{2}}$, we get,
\[9-\dfrac{10{{(a-d)}^{2}}}{\beta }-\dfrac{20}{{{\beta }^{2}}}=0\]
\[9-10{{(a-d)}^{2}}\left( \dfrac{1}{\beta } \right)-20\left( \dfrac{1}{{{\beta }^{2}}} \right)=0\]
\[20\left( \dfrac{1}{{{\beta }^{2}}} \right)+10{{(a-d)}^{2}}\left( \dfrac{1}{\beta } \right)-9=0\]
Comparing the above equation with $20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0$we can clearly see that $\dfrac{1}{\beta }$ is also a root of $20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0$.
Hence, from above statements, roots of the equations $\left( \dfrac{1}{u}+\dfrac{1}{v}+\dfrac{1}{w} \right){{x}^{2}}+\left[ {{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(d-b)}^{2}} \right]x+u+v+w=0$ (which can be simplified as $9{{x}^{2}}-10{{(a-d)}^{2}}x-20=0$ ) and $20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0$are reciprocals.
$\therefore $ The correct answer is an option (c).

Note:

 $u,v$ and$w$ in equations (1), (2), and (3) can also be solved by using matrix algebra.
Solve $X={{A}^{-1}}B$, where, matrix of variables\[X=\left( \begin{matrix}
   u \\
   v \\
   w \\
\end{matrix} \right)\], matrix of coefficients $A=\left( \begin{matrix}
   1 & 2 & 3 \\
   4 & 5 & 6 \\
   6 & 9 & 0 \\
\end{matrix} \right)$,
 matrix of constants $B=\left( \begin{matrix}
   6 \\
   12 \\
   4 \\
\end{matrix} \right)$ .
Finding the roots of the quadratic equations $9{{x}^{2}}-10{{(a-d)}^{2}}x-20=0$ and $20{{x}^{2}}+10{{(a-d)}^{2}}x-9=0$ separately and then comparing the roots is also an alternative, but tedious method.
Substituting the properties of real numbers a, b, c, d in G.P , such as ,${{b}^{2}}=ac$and ${{c}^{2}}=bd$,$bc=ad$ , has to be done as needed. Inappropriate substitutions might lead to more complexities.