Question

Solve the equation $$40x^4-22x^3-21x^2+2x+1=0$$, 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 $${\displaystyle \frac{1}{a-3d}},{\displaystyle \frac{1}{a-d}},{\displaystyle \frac{1}{a+d}}\text{ and }{\displaystyle \frac{1}{a+3d}}$$. Use this concept to reach the solution of the problem.

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

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

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.