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Solution

given 3x²(x²+2)=20-x²

3x²(x²+2)=22-(x²+2)

  3x²(x²+2) + (x²+2)=22

let t=x²+2

3(t-2)t + t - 22=0

3t² - 6t + t - 22=0

3t² - 5t - 22=0

t = [ 5 +/- sqrt(25+264) ]/6

= [ 5 +/- 17 ]/6

t = 22/6 and t=-12/6

t = 11/3, t = -2

therefore x²+2 = 11/3 and x²+2=-2

x² = 5/3 and x² = -4

x = sqrt(5/3) ,-sqrt(5/3) , 2i