find the positive root of 3xsqrare +6= 9
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3 x^2 +6= 9
3x^+6 - 9 = 0
3(x^2 +2-3 ) = 0
x^2 -1=0
x^2 = 1
x = root 1
so (1) is the positive root for this equation
3x^+6 - 9 = 0
3(x^2 +2-3 ) = 0
x^2 -1=0
x^2 = 1
x = root 1
so (1) is the positive root for this equation
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0
3x^2 +6 = 9
3x^2 = 9 - 6
3x^2 = 3
x*x = 1 *1
x = +1, -1
3x^2 = 9 - 6
3x^2 = 3
x*x = 1 *1
x = +1, -1
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