What is the answer to 2u^{2/3}+7u^{1/3}=4 ?

The answer to 2u^{2/3}+7u^{1/3}=4 is u= 1/8 ,u=-64

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2u^{2/3}+7u^{1/3}=4

Solution

2 u^{\frac{2}{3}} + 7 u^{\frac{1}{3}} = 4

Solution

u = \frac{1}{8}, u = - 64

Decimal

u = 0.125, u = - 64

Solution steps

2 u^{\frac{2}{3}} + 7 u^{\frac{1}{3}} = 4

Use the following exponent property

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

u^{\frac{2}{3}} = (u^{\frac{1}{3}})^{2}

2 (u^{\frac{1}{3}})^{2} + 7 u^{\frac{1}{3}} = 4

Rewrite the equation with

u^{\frac{1}{3}} = v

2 v^{2} + 7 v = 4

Solve

2 v^{2} + 7 v = 4 : v = \frac{1}{2}, v = - 4

v = \frac{1}{2}, v = - 4

Substitute back

v = u^{\frac{1}{3}},

solve for

u

Solve

u^{\frac{1}{3}} = \frac{1}{2} : u = \frac{1}{8}

Solve

u^{\frac{1}{3}} = - 4 : u = - 64

u = \frac{1}{8}, u = - 64

Verify Solutions:

u = \frac{1}{8}

True

, u = - 64

True

The solutions are

u = \frac{1}{8}, u = - 64

Graph

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solvefor x, u/(sqrt(u^2+1))=x solve for

x, \frac{u}{u ^{2} + 1} = x

x+2=sqrt(2x+7)

x + 2 = 2 x + 7

sqrt(x+1)+1=0

x + 1 + 1 = 0

1= 1/(3x^{2/3)}

1 = \frac{1}{3 x ^{\frac{2}{3}}}

sqrt(x)-1=sqrt(x-3)

x - 1 = x - 3

Frequently Asked Questions (FAQ)

What is the answer to 2u^{2/3}+7u^{1/3}=4 ?
The answer to 2u^{2/3}+7u^{1/3}=4 is u= 1/8 ,u=-64
Solve for u: 2u^{2/3}+7u^{1/3}=4
The solution is u= 1/8 ,u=-64

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2u^{2/3}+7u^{1/3}=4

Solution

Solution

Graph

Popular Examples

Frequently Asked Questions (FAQ)

What is the answer to 2u^{2/3}+7u^{1/3}=4 ?

Solve for u: 2u^{2/3}+7u^{1/3}=4

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