11a. x^(2/3)+6x^(1/3)=7 As we have here rational degrees, x must be non-negative:x>=0 Make a substitution y = x^(1/3)> 0 Then y^2 + 6y - 7 =0 D = 36 + 4*7 =64 = 8^2 y1 = (-6+8)/2 =1 y2 = (-6-8)/2 =-7 < 0
Since y>0, we obtain only one solution y1 = 1 =x^(1/3) whence x=1
Answer: 1 ---------------------------------
11b. (y+3)^2-8(y+3)+12=0
Make a substitution t = y+3 Then t^2 - 8t + 12 =0 D = 64 - 48 =16 = 4^2 t1 = (8+4)/2 =6 t2 = (8-4)/2 =2
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