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Solve the equation 40x422x321x2+2x+1=0, the roots of which are in harmonic progression.

Answer
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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 1a3d,1ad,1a+d and 1a+3d. Use this concept to reach the solution of the problem.

Complete step-by-step answer:
Given equation is 40x422x321x2+2x+1=0
Let the roots of this equation are 1a3d,1ad,1a+d and 1a+3d.
Transforming the equation by replacing x with 1x.
40(1x)422(1x)321(1x)2+2(1x)+1=040x422x321x2+2x+1=04022x21x2+2x3+x4x4=0x4+2x321x222x40=0

Hence the transformed equation is x4+2x321x222x40=0 and the roots are transformed into a3d,ad,a+d and a + 3d.
We know that for the equation ax4+bx3+cx2+dx+e=0, the sum of the roots S1=ba and the sum of the roots taken two at a time is S2=ca.
So, by using the above formula for the equation x4+2x321x222x40=0 we have
Sum of the roots S1=ba=21
S1=(a3d)+(ad)+(a+d)+(a+3d)=21S1=4a=24a=2a=12
Sum of the roots taken two at a time S2=ca=211=21
S2=(a3d)(ad)+(a3d)(a+d)+(a3d)(a+3d)+(ad)(a+d)+(ad)(a+3d)+(a+d)(a+3d)=21S2=a24d+3d2+a22d3d2+a29d2+a2d2+a2+2d3d2+a2+4d+3d2=21S2=6a210d2=21
Since, a=12
S2=6(12)210d2=2110d2=6(14)+2110d2=6+844d2=904×110d2=9040=94d=32
So, the roots are
1a3d=1123(32)=1(1+9)2=1102=151ad=11232=142=121a+d=112+32=22=11a+3d=112+3(32)=1(1+9)2=28=14
Hence the roots of the equation 40x422x321x2+2x+1=0 are 15,12,1 and 14

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.
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