Question 352103
My problem is finding all real solutions with the following equation {{{x^(4/3)-5x^(2/3)+6}}}
I put 6 on the other side so it turns into
{{{x^(4/3)-5x^(2/3)=-6}}}
I can't combine the two terms so I ended up subtracting -5x^(2/3) from both sides to get:
{{{x^(4/3)=5x^(2/3)-6}}}
Then I will take the power 3/4th to both sides so X on the left is by itself.
{{{x=(5x^(2/3))^(3/4)-6^(3/4)}}}
Which will simplify to:
{{{125^(1/4)*x^(1/2)-126^(1/4)=X}}}
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x^(4/3)-5x^(2/3)+6}
Your problem is a quadratic with variable x^(2/3).
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It can be written as:
[x^(2/3)]^2 - 5[x^(2/3)] +6 = 0
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It factors as:
[x^(2/3)-3][x^(2/3)-2] = 0
Then:
x^(2/3) = 3 or x^(2/3) = 2
x = 3^(3/2) or x = 1^(3/2)
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x = 3sqrt(3) or x = 1
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Cheers,
Stan H.