Solve for x square root of x*(x+2)+4=2

$\sqrt{x \cdot (x + 2) + 4} = 2$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x \cdot (x + 2) + 4}}^{2} = 2^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x \cdot (x + 2) + 4}$ as ${(x \cdot (x + 2) + 4)}^{\frac{1}{2}}$ .

${({(x \cdot (x + 2) + 4)}^{\frac{1}{2}})}^{2} = 2^{2}$

Simplify the left side.

Simplify ${({(x \cdot (x + 2) + 4)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x \cdot (x + 2) + 4)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x \cdot (x + 2) + 4)}^{\frac{1}{2} \cdot 2} = 2^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x \cdot (x + 2) + 4)}^{\frac{1}{2} \cdot 2} = 2^{2}$

Rewrite the expression.

${(x \cdot (x + 2) + 4)}^{1} = 2^{2}$

Simplify each term.

Apply the distributive property.

${(x \cdot x + x \cdot 2 + 4)}^{1} = 2^{2}$

Multiply $x$ by $x$ .

${(x^{2} + x \cdot 2 + 4)}^{1} = 2^{2}$

Move $2$ to the left of $x$ .

${(x^{2} + 2 x + 4)}^{1} = 2^{2}$

Simplify.

$x^{2} + 2 x + 4 = 2^{2}$

Simplify the right side.

Raise $2$ to the power of $2$ .

$x^{2} + 2 x + 4 = 4$

Solve for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} + 2 x + 4 - 4 = 0$

Combine the opposite terms in $x^{2} + 2 x + 4 - 4$ .

Subtract $4$ from $4$ .

$x^{2} + 2 x + 0 = 0$

Add $x^{2} + 2 x$ and $0$ .

$x^{2} + 2 x = 0$

Factor $x$ out of $x^{2} + 2 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x + 2 x = 0$

Factor $x$ out of $2 x$ .

$x \cdot x + x \cdot 2 = 0$

Factor $x$ out of $x \cdot x + x \cdot 2$ .

$x (x + 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $x (x + 2) = 0$ true.

$x = 0, - 2$

Solve for x square root of x*(x+2)+4=2

Related Questions

Support LCMGCF.com by disabling your adblocker.