Solve for y

y=\sqrt{x\left(x^{2}-4\right)}<br/>y=-\sqrt{x\left(x^{2}-4\right)}\text{, }\left(x\geq -2\text{ and }x\leq 0\right)\text{ or }x\geq 2

$y=x(x_{2}−4) $

$y=−x(x_{2}−4) ,(x≥−2andx≤0)orx≥2$

$y=−x(x_{2}−4) ,(x≥−2andx≤0)orx≥2$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Solve for x (complex solution)

x=-\frac{3\times \left(\frac{2}{9}\right)^{\frac{2}{3}}\left(1+\sqrt{3}i\right)\left(\sqrt{3\left(27y^{4}-256\right)}+9y^{2}\right)^{-\frac{1}{3}}\left(3\left(-\sqrt{3}i-1\right)\times \left(\frac{\sqrt{3\left(27y^{4}-256\right)}+9y^{2}}{9}\right)^{\frac{2}{3}}+8\times 2^{\frac{2}{3}}\right)}{8}<br/>x=\frac{3\times \left(\frac{2}{9}\right)^{\frac{2}{3}}\left(\sqrt{3\left(27y^{4}-256\right)}+9y^{2}\right)^{-\frac{1}{3}}\left(3\times \left(\frac{\sqrt{3\left(27y^{4}-256\right)}+9y^{2}}{9}\right)^{\frac{2}{3}}+4\times 2^{\frac{2}{3}}\right)}{2}<br/>x=-\frac{3\times \left(\frac{2}{9}\right)^{\frac{2}{3}}\left(-\sqrt{3}i+1\right)\left(\sqrt{3\left(27y^{4}-256\right)}+9y^{2}\right)^{-\frac{1}{3}}\left(3\left(-1+\sqrt{3}i\right)\times \left(\frac{\sqrt{3\left(27y^{4}-256\right)}+9y^{2}}{9}\right)^{\frac{2}{3}}+8\times 2^{\frac{2}{3}}\right)}{8}

$x=−83×(92 )_{32}(1+3 i)(3(27y_{4}−256) +9y_{2})_{−31}(3(−3 i−1)×(93(27y−256) +9y )_{32}+8×2_{32}) $

$x=23×(92 )_{32}(3(27y_{4}−256) +9y_{2})_{−31}(3×(93(27y−256) +9y )_{32}+4×2_{32}) $

$x=−83×(92 )_{32}(−3 i+1)(3(27y_{4}−256) +9y_{2})_{−31}(3(−1+3 i)×(93(27y−256) +9y )_{32}+8×2_{32}) $

$x=23×(92 )_{32}(3(27y_{4}−256) +9y_{2})_{−31}(3×(93(27y−256) +9y )_{32}+4×2_{32}) $

$x=−83×(92 )_{32}(−3 i+1)(3(27y_{4}−256) +9y_{2})_{−31}(3(−1+3 i)×(93(27y−256) +9y )_{32}+8×2_{32}) $

Solve for y (complex solution)

y=-\sqrt{x\left(x^{2}-4\right)}<br/>y=\sqrt{x\left(x^{2}-4\right)}

$y=−x(x_{2}−4) $

$y=x(x_{2}−4) $

$y=x(x_{2}−4) $

Graph

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y=\sqrt{x^{3}-4x} y=-\sqrt{x^{3}-4x}

Take the square root of both sides of the equation.

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

Subtract x^{3} from both sides.

y^{2}-x^{3}+4x=0

Add 4x to both sides.

y^{2}+4x-x^{3}=0

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

y=\frac{0±\sqrt{0^{2}-4\left(4x-x^{3}\right)}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and 4x-x^{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

y=\frac{0±\sqrt{-4\left(4x-x^{3}\right)}}{2}

Square 0.

y=\frac{0±\sqrt{4x^{3}-16x}}{2}

Multiply -4 times 4x-x^{3}.

y=\frac{0±2\sqrt{x^{3}-4x}}{2}

Take the square root of -16x+4x^{3}.

y=\sqrt{x^{3}-4x}

Now solve the equation y=\frac{0±2\sqrt{x^{3}-4x}}{2} when ± is plus.

y=-\sqrt{x^{3}-4x}

Now solve the equation y=\frac{0±2\sqrt{x^{3}-4x}}{2} when ± is minus.

y=\sqrt{x^{3}-4x} y=-\sqrt{x^{3}-4x}

The equation is now solved.

Examples

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$x_{2}−4x−5=0$

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4 \sin \theta \cos \theta = 2 \sin \theta

$4sinθcosθ=2sinθ$

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y = 3x + 4

$y=3x+4$

Arithmetic

699 * 533

$699∗533$

Matrix

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \right]

$[25 34 ][2−1 01 35 ]$

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

${8x+2y=467x+3y=47 $

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

$dxd (x−5)(3x_{2}−2) $

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

$∫_{0}xe_{−x_{2}}dx$

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

$x→−3lim x_{2}+2x−3x_{2}−9 $