Answer

Verified

482.4k+ views

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.

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.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers