What is the answer to solvefor y,x^{2/3}+y^{2/3}=4 ?

The answer to solvefor y,x^{2/3}+y^{2/3}=4 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,x^{2/3}+y^{2/3}=4

solve for

y, x^{\frac{2}{3}} + y^{\frac{2}{3}} = 4

y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}

Solution steps

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 4

Move

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

to the right side

y^{\frac{2}{3}} = 4 - x^{\frac{2}{3}}

Take both sides of the equation to the power of

3 : y^{2} = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

y^{2} = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

Solve

y^{2} = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} : y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

Verify Solutions:

y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}

The solutions are

y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}

(x) = - 1

8 x + 9 = x

x + 6 - 5 x + 1 + 1 = 0

x + 2 = 4 x - 2

\frac{x}{1024} \cdot 72 = 367

What is the answer to solvefor y,x^{2/3}+y^{2/3}=4 ?
The answer to solvefor y,x^{2/3}+y^{2/3}=4 is