Question

Solve the equation \[40{x^4} - 22{x^3} - 21{x^2} + 2x + 1 = 0\], the roots of which are in harmonic progression.

Answer 1

Hint: In the given question the roots of the equation are in harmonic progression (H.P.). So, take the roots as \[\dfrac{1}{{a - 3d}},\dfrac{1}{{a - d}},\dfrac{1}{{a + d}}{\text{ and }}\dfrac{1}{{a + 3d}}\]. Use this concept to reach the solution of the problem.

Complete step-by-step answer:
Given equation is \[40{x^4} - 22{x^3} - 21{x^2} + 2x + 1 = 0\]
Let the roots of this equation are \[\dfrac{1}{{a - 3d}},\dfrac{1}{{a - d}},\dfrac{1}{{a + d}}{\text{ and }}\dfrac{1}{{a + 3d}}\].
Transforming the equation by replacing \[x\] with \[\dfrac{1}{x}\].
\[
   \Rightarrow 40{\left( {\dfrac{1}{x}} \right)^4} - 22{\left( {\dfrac{1}{x}} \right)^3} - 21{\left( {\dfrac{1}{x}} \right)^2} + 2\left( {\dfrac{1}{x}} \right) + 1 = 0 \\
   \Rightarrow \dfrac{{40}}{{{x^4}}} - \dfrac{{22}}{{{x^3}}} - \dfrac{{21}}{{{x^2}}} + \dfrac{2}{x} + 1 = 0 \\
   \Rightarrow \dfrac{{40 - 22x - 21{x^2} + 2{x^3} + {x^4}}}{{{x^4}}} = 0 \\
   \Rightarrow {x^4} + 2{x^3} - 21{x^2} - 22x - 40 = 0 \\
\]

Hence the transformed equation is \[{x^4} + 2{x^3} - 21{x^2} - 22x - 40 = 0\] and the roots are transformed into \[a - 3d,a - d,a + d{\text{ and a + 3d}}\].
We know that for the equation \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\], the sum of the roots \[{S_1} = - \dfrac{b}{a}\] and the sum of the roots taken two at a time is \[{S_2} = \dfrac{c}{a}\].
So, by using the above formula for the equation \[{x^4} + 2{x^3} - 21{x^2} - 22x - 40 = 0\] we have
Sum of the roots \[{S_1} = - \dfrac{b}{a} = - \dfrac{2}{1}\]
\[
   \Rightarrow {S_1} = (a - 3d) + (a - d) + (a + d) + \left( {a + 3d} \right) = - \dfrac{2}{1} \\
   \Rightarrow {S_1} = 4a = - 2 \\
   \Rightarrow 4a = - 2 \\
  \therefore a = - \dfrac{1}{2} \\
\]
Sum of the roots taken two at a time \[{S_2} = \dfrac{c}{a} = \dfrac{{ - 21}}{1} = - 21\]
\[
   \Rightarrow {S_2} = (a - 3d)(a - d) + (a - 3d)\left( {a + d} \right) + \left( {a - 3d} \right)\left( {a + 3d} \right) + \left( {a - d} \right)\left( {a + d} \right) + \left( {a - d} \right)\left( {a + 3d} \right) + \left( {a + d} \right)\left( {a + 3d} \right) = - 21 \\
   \Rightarrow {S_2} = {a^2} - 4d + 3{d^2} + {a^2} - 2d - 3{d^2} + {a^2} - 9{d^2} + {a^2} - {d^2} + {a^2} + 2d - 3{d^2} + {a^2} + 4d + 3{d^2} = - 21 \\
   \Rightarrow {S_2} = 6{a^2} - 10{d^2} = - 21 \\
\]
Since, \[a = - \dfrac{1}{2}\]
\[
  {S_2} = 6{\left( { - \dfrac{1}{2}} \right)^2} - 10{d^2} = - 21 \\
  10{d^2} = 6\left( {\dfrac{1}{4}} \right) + 21 \\
  10{d^2} = \dfrac{{6 + 84}}{4} \\
  {d^2} = \dfrac{{90}}{4} \times \dfrac{1}{{10}} \\
  {d^2} = \dfrac{{90}}{{40}} = \dfrac{9}{4} \\
  \therefore d = \dfrac{3}{2} \\
\]
So, the roots are
\[
   \Rightarrow \dfrac{1}{{a - 3d}} = \dfrac{1}{{ - \dfrac{1}{2} - 3\left( {\dfrac{3}{2}} \right)}} = \dfrac{1}{{ - \dfrac{{(1 + 9)}}{2}}} = \dfrac{1}{{ - \dfrac{{10}}{2}}} = - \dfrac{1}{5} \\
   \Rightarrow \dfrac{1}{{a - d}} = \dfrac{1}{{ - \dfrac{1}{2} - \dfrac{3}{2}}} = \dfrac{1}{{ - \dfrac{4}{2}}} = - \dfrac{1}{2} \\
   \Rightarrow \dfrac{1}{{a + d}} = \dfrac{1}{{ - \dfrac{1}{2} + \dfrac{3}{2}}} = \dfrac{2}{2} = 1 \\
   \Rightarrow \dfrac{1}{{a + 3d}} = \dfrac{1}{{ - \dfrac{1}{2} + 3\left( {\dfrac{3}{2}} \right)}} = \dfrac{1}{{\dfrac{{( - 1 + 9)}}{2}}} = \dfrac{2}{8} = \dfrac{1}{4} \\
\]
Hence the roots of the equation \[40{x^4} - 22{x^3} - 21{x^2} + 2x + 1 = 0\] are \[ - \dfrac{1}{5}, - \dfrac{1}{2},1{\text{ and }}\dfrac{1}{4}\]

Note: The given equation is bi-quadratic equation. Hence the equation has 4 roots. Whenever the equation is transformed, the roots of that equation will also be transformed accordingly.