Solveeit Logo
Login

Question

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

Solve the equation 40x4−22x3−21x2+2x+1=040{x^4} - 22{x^3} - 21{x^2} + 2x + 1 = 0, the roots of which are in harmonic progression.

Explanation

Solution

Hint: In the given question the roots of the equation are in harmonic progression (H.P.). So, take the roots as 1a−3d,1a−d,1a+d and 1a+3d\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 40x4−22x3−21x2+2x+1=040{x^4} - 22{x^3} - 21{x^2} + 2x + 1 = 0
Let the roots of this equation are 1a−3d,1a−d,1a+d and 1a+3d\dfrac{1}{{a - 3d}},\dfrac{1}{{a - d}},\dfrac{1}{{a + d}}{\text{ and }}\dfrac{1}{{a + 3d}}.
Transforming the equation by replacing xx with 1x\dfrac{1}{x}.

⇒40(1x)4−22(1x)3−21(1x)2+2(1x)+1=0 ⇒40x4−22x3−21x2+2x+1=0 ⇒40−22x−21x2+2x3+x4x4=0 ⇒x4+2x3−21x2−22x−40=0  \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 x4+2x3−21x2−22x−40=0{x^4} + 2{x^3} - 21{x^2} - 22x - 40 = 0 and the roots are transformed into a−3d,a−d,a+d and a + 3da - 3d,a - d,a + d{\text{ and a + 3d}}.
We know that for the equation ax4+bx3+cx2+dx+e=0a{x^4} + b{x^3} + c{x^2} + dx + e = 0, the sum of the roots S1=−ba{S_1} = - \dfrac{b}{a} and the sum of the roots taken two at a time is S2=ca{S_2} = \dfrac{c}{a}.
So, by using the above formula for the equation x4+2x3−21x2−22x−40=0{x^4} + 2{x^3} - 21{x^2} - 22x - 40 = 0 we have
Sum of the roots S1=−ba=−21{S_1} = - \dfrac{b}{a} = - \dfrac{2}{1}

⇒S1=(a−3d)+(a−d)+(a+d)+(a+3d)=−21 ⇒S1=4a=−2 ⇒4a=−2 ∴a=−12  \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 S2=ca=−211=−21{S_2} = \dfrac{c}{a} = \dfrac{{ - 21}}{1} = - 21

⇒S2=(a−3d)(a−d)+(a−3d)(a+d)+(a−3d)(a+3d)+(a−d)(a+d)+(a−d)(a+3d)+(a+d)(a+3d)=−21 ⇒S2=a2−4d+3d2+a2−2d−3d2+a2−9d2+a2−d2+a2+2d−3d2+a2+4d+3d2=−21 ⇒S2=6a2−10d2=−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=−12a = - \dfrac{1}{2}

S2=6(−12)2−10d2=−21 10d2=6(14)+21 10d2=6+844 d2=904×110 d2=9040=94 ∴d=32  {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

⇒1a−3d=1−12−3(32)=1−(1+9)2=1−102=−15 ⇒1a−d=1−12−32=1−42=−12 ⇒1a+d=1−12+32=22=1 ⇒1a+3d=1−12+3(32)=1(−1+9)2=28=14  \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 40x4−22x3−21x2+2x+1=040{x^4} - 22{x^3} - 21{x^2} + 2x + 1 = 0 are −15,−12,1 and 14 - \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.