What is the answer to 6u^{-1/2}+13u^{-1/4}+6=0 ?

The answer to 6u^{-1/2}+13u^{-1/4}+6=0 is No Solution for u\in\mathbb{R}

Solve for u: 6u^{-1/2}+13u^{-1/4}+6=0

The solution is No Solution for u\in\mathbb{R}

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6 u^{- \frac{1}{2}} + 13 u^{- \frac{1}{4}} + 6 = 0

No Solution for u \in R

Solution steps

6 u^{- \frac{1}{2}} + 13 u^{- \frac{1}{4}} + 6 = 0

Expand

6 u^{- \frac{1}{2}} + 13 u^{- \frac{1}{4}} + 6 : \frac{6}{u ^{\frac{1}{2}}} + \frac{13}{u ^{\frac{1}{4}}} + 6

\frac{6}{u ^{\frac{1}{2}}} + \frac{13}{u ^{\frac{1}{4}}} + 6 = 0

Use the following exponent property

: a^{\frac{1}{n}} = (a^{\frac{1}{m}})^{\frac{m}{n}}

u^{\frac{1}{2}} = (u^{\frac{1}{4}})^{\frac{4}{2}}

\frac{6}{( u ^{\frac{1}{4}} ) ^{2}} + \frac{13}{u ^{\frac{1}{4}}} + 6 = 0

Rewrite the equation with

u^{\frac{1}{4}} = v

\frac{6}{v ^{2}} + \frac{13}{v} + 6 = 0

Solve

\frac{6}{v ^{2}} + \frac{13}{v} + 6 = 0 : v = - \frac{2}{3}, v = - \frac{3}{2}

v = - \frac{2}{3}, v = - \frac{3}{2}

Substitute back

v = u^{\frac{1}{4}},

solve for

u

Solve

u^{\frac{1}{4}} = - \frac{2}{3} :

No Solution for

u \in R

Solve

u^{\frac{1}{4}} = - \frac{3}{2} :

No Solution for

u \in R

No Solution for u \in R

17 - x = x - 5

\frac{1 - x}{1 + x} = 1 - x

4 x + 8 = 33 x

x + 3 x - 12 = 0

x + 6 - 2 x - 4 = 0

What is the answer to 6u^{-1/2}+13u^{-1/4}+6=0 ?
The answer to 6u^{-1/2}+13u^{-1/4}+6=0 is No Solution for u\in\mathbb{R}
Solve for u: 6u^{-1/2}+13u^{-1/4}+6=0
The solution is No Solution for u\in\mathbb{R}