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

Last updated date: 19th Jul 2024
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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.

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.