What is y=sqrt(2x^2+z^2),y=6-2x^2-z^2 ?

The solution to y=sqrt(2x^2+z^2),y=6-2x^2-z^2 is

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Popular >

y=sqrt(2x^2+z^2),y=6-2x^2-z^2

Solution

y = 2 x^{2} + z^{2}, y = 6 - 2 x^{2} - z^{2}

Solution

x \in R, x \in R, x \in R, x \in R, y = 2, y = 2, y = 2, y = 2, z = 4 - 2 x^{2} z = - 4 - 2 x^{2} z = 4 - 2 x^{2} z = - 4 - 2 x^{2}

Solution steps

[y = 2 x^{2} + z^{2} y = 6 - 2 x^{2} - z^{2}]

Isolate

z

for

y = z^{2} + 2 x^{2} : z = y^{2} - 2 x^{2}, z = - y^{2} - 2 x^{2}

Plug the solutions

z = y^{2} - 2 x^{2}, z = - y^{2} - 2 x^{2}

into

y = 6 - 2 x^{2} - z^{2}

For

y = 6 - 2 x^{2} - z^{2}

, subsitute

z

with

y^{2} - 2 x^{2} : y = 2 or y = - 3

For

y = 6 - 2 x^{2} - z^{2}

, subsitute

z

with

- y^{2} - 2 x^{2} : y = 2 or y = - 3

Solve

y = z^{2} + 2 x^{2}

for

y = 2 : (x \in R, y = 2, z = 4 - 2 x^{2}), (x \in R, y = 2, z = - 4 - 2 x^{2})

Solve

y = z^{2} + 2 x^{2}

for

y = - 3 :

No Solution

Solve

y = z^{2} + 2 x^{2}

for

y = 2 : (x \in R, y = 2, z = 4 - 2 x^{2}), (x \in R, y = 2, z = - 4 - 2 x^{2})

Solve

y = z^{2} + 2 x^{2}

for

y = - 3 :

No Solution

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

y = 2 x^{2} + z^{2}, y = 6 - 2 x^{2} - z^{2}

are

x \in R, x \in R, x \in R, x \in R, y = 2, y = 2, y = 2, y = 2, z = 4 - 2 x^{2} z = - 4 - 2 x^{2} z = 4 - 2 x^{2} z = - 4 - 2 x^{2}

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a+b=27,a*b=182

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Frequently Asked Questions (FAQ)

What is y=sqrt(2x^2+z^2),y=6-2x^2-z^2 ?
The solution to y=sqrt(2x^2+z^2),y=6-2x^2-z^2 is