Question

xp+3yp=2(z-x2p2) solve this equation.

Answer 1

Answer on Question #50307 – Math - Algebra

xp+3yp=2(z-x2p2) solve this equation.

Solution

$$x p + y p = 2 (z - x ^ {2} p ^ {2})$$

Solve for $x$

(other values of $y, z, p$ should be known in order to get a definite solution):

$2p^{2}x^{2} + px + (yp - 2z) = 0$ is a quadratic equation with respect to $x$, when $p \neq 0$;

its discriminant is $D1 = p^2 - 4 \cdot 2p^2(yp - 2z) = p^2 - 8yp^3 + 16zp^2$

Solutions are the following:

If $p \neq 0$, then

$$x = \frac {- p \pm \sqrt {D 1}}{2 \cdot 2 p ^ {2}} = \frac {- p \pm p \sqrt {1 - 8 y p + 1 6 z}}{4 p ^ {2}} = - \frac {1}{4 p} - \frac {1}{4 p} \sqrt {1 - 8 y p + 1 6 z}; - \frac {1}{4 p} - \frac {1}{4 p} \sqrt {1 - 8 y p + 1 6 z}.$$

If $p = 0$, then $0 = 2z$, hence

$x = C$ is an arbitrary real number, $y = E$ is an arbitrary real number, $z = 0$.

Solve for $y$

(other values of $x, z, p$ should be known in order to get a definite solution):

$xp + yp = 2(z - x^2 p^2)$ is a linear equation with respect to $y$.

If $p \neq 0$, then $xp + yp = 2(z - x^2 p^2) \Rightarrow |subtract xp \text{ from both sides}| \Rightarrow$

$$\Rightarrow y p = 2 \left(z - x ^ {2} p ^ {2}\right) - x p \Rightarrow | \text{divide both sides by } p | \Rightarrow$$ $$y = \frac {2 (z - x ^ {2} p ^ {2}) - x p}{p} = \frac {2 z}{p} - 2 x ^ {2} p - x$$

If $p = 0$, then $0 = 2z$, hence

$y = E$ is an arbitrary real number, $x = C$ is an arbitrary real number, $z = 0$.

Solve for $z$

(other values of $x, y, p$ should be known in order to get a definite solution):

$xp + yp = 2(z - x^2 p^2)$ is a linear equation with respect to $z$;

$\Rightarrow |add2x^{2}p^{2}$ to both sides| $\Rightarrow xp + yp + 2x^{2}p^{2} = 2z \Rightarrow |divide both sides by 2| \Rightarrow$

$$\frac {x p + y p + 2 x ^ {2} p ^ {2}}{2} = z, \text{ i.e. } z = \frac {x p + y p + 2 x ^ {2} p ^ {2}}{2}$$

Solve for $p$

(other values of $x, y, z$ should be known in order to get a definite solution):

$2x^{2}p^{2} + (x + y)p - 2z = 0$ is a quadratic equation with respect to $p$, when $x \neq 0$;

its discriminant is $D2 = (x + y)^2 - 4 \cdot 2x^2 \cdot (-2z) = x^2 + y^2 + 2xy + 16x^2z$

Solutions are the following:

If $x \neq 0$, then

$$p = \frac {-(x + y) \pm \sqrt {D 2}}{2 \cdot 2 x ^ {2}} = \frac {-(x + y) \pm \sqrt {x ^ {2} + y ^ {2} + 2 x y + 1 6 x ^ {2} z}}{4 x ^ {2}} = \frac {-(x + y) - \sqrt {x ^ {2} + y ^ {2} + 2 x y + 1 6 x ^ {2} z}}{4 x ^ {2}}; \frac {-(x + y) + \sqrt {x ^ {2} + y ^ {2} + 2 x y + 1 6 x ^ {2} z}}{4 x ^ {2}}.$$

If $x = 0$, then $yp = 2z$ is a linear equation with respect to $p$, hence

If $x = 0$ and $y \neq 0$, then $p = \frac{2z}{y}$;

If $x = 0$ and $y = 0$ and $z \neq 0$, then the equation does not have solutions

If $x = 0$ and $y = 0$ and $z = 0$, then the equation has infinitely many solutions, i.e. $p = K$, where $K$ is an arbitrary real constant.

Note that there are infinitely many solutions when two or less values of $x, y, z, p$ are known, they should satisfy condition $z \geq \frac{xp + yp}{2}$.

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