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`        find the solution of the equation 12x4-56x3+89x2-56x+12=0i challenge you to find all the 4 real roots`
7 months ago

Arun
22998 Points
```							Note the symmetric coefficients. x=0 is not a root so start by dividing by x^2. This gives. 12(x^2 + 1/x^2) - 56(x+1/x)+89 = 0 Now subtract 12(x+1/x)^2 from both sides and simplify - 56(x+1/x)+65 = -12(x+1/x)^2 Let u=x+1/x and you now have the quadratic equation 12u^2 - 56u + 65 = 0 (2u-5)(6u-13) = 0 u=5/2 or u=13/6 x+1/x=5/2 gives x=2 or x=1/2 and x+1/x = 13/6 gives x=3/2 or x=2/3.
```
7 months ago
ritika
21 Points
```							12x^4 – 56x^3 + 89x^2 – 56x + 12 = 0Dividing by x^2;12x^2 – 56x + 89 – 56/x + 12/x^2 = 0(12x^2 + 12/x^2) – (56x + 56/x) + 89 = 012(x^2 + 1/x^2) – 56(x + 1/x) + 89 = 012[(x + 1/x)^2 – 2] – 56(x + 1/x) + 89 = 0Let m = x + 1/x   …........(1)12(m^2 – 2) – 56m + 89 = 012m^2 – 56m + 89 – 24 = 012m^2 – 56m + 65 = 0 12m^2 – 30m – 26m + 65 = 06m(2m – 5) – 13(2m – 5) = 0(2m – 5)(6m – 13) = 02m – 5 = 0 and 6m – 13 = 0m = 5/2 and m = 13/6 From (1);x + 1/x = 5/2 and x + 1/x = 13/62x^2 – 5x + 2 = 0 and 6x^2 – 13x + 6 = 02x^2 – 4x – x + 2 = 0 and 6x^2 – 9x – 4x + 6 = 0(x – 2)(2x – 1) = 0 and (2x – 3)(3x – 2) = 0x = 2, x = ½ and x = 3/2, x = 2/3
```
7 months ago
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