Question

Let $a,b,c,d$ be real number 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 $(\frac{1}{u}+\frac{1}{v}+\frac{1}{w})x^2+[{(b-c)}^2+{(c-a)}^2+{(d-b)}^2]x+u+v+w=0$ and ${20x}^2+10{(a-d)}^{2}x-9=0$ are

• A. equal
• B. imaginary
• C. reciprocals
• D. none of these

Answer 1

Answer: (C)

reciprocals

Solving for $u,v,w$, we get

$u=-\frac{1}{3},v=\frac{2}{3}$ and $w=\frac{5}{3}$

Now,

${(b-c)}^2+{(c-a)}^2+{(d-b)}^2=a^2+2b^2+2c^2+d^2-2bc-2ca-2bd$

As $a,b,c,d$ are in G.P $bc=ad$, $ca=b^2$ and $bd=c^2$

Therefore ${(b-c)}^2+{(c-a)}^2+{(d-b)}^2={(a-d)}^2$

The quadratic equation

$(\frac{1}{u}+\frac{1}{v}+\frac{1}{w})x^2+({(b-c)}^2+{(c-a)}^2+{(d-b)}^2)x+u+v+w=0\Rightarrow -\frac{9}{10}{x}^2+{(a-d)}^{2}x+2=0$

$\Rightarrow 9x^2-10{(a-d)}^{2}x-20=0$ ...(1)

Equation whose roots are reciprocal of (1) is

$9{\left(\frac{1}{x}\right)}^2-10{(a-d)}^2\left(\frac{1}{x}\right)-20=0\Rightarrow 9-10{(a-d)}^{2}x-20x^2=0$

i.e. $20x^2+10{(a-d)}^{2}x-9=0$