Solve for x square root of x^2+x-2=2

$\sqrt{x^{2} + x - 2} = 2$

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

${\sqrt{x^{2} + x - 2}}^{2} = 2^{2}$

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

Solve for $x$ .

Subtract $4$ from both sides of the equation.

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

Subtract $4$ from $- 2$ .

$x^{2} + x - 6 = 0$

Factor $x^{2} + x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

$(x - 2) (x + 3) = 0$

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

$x - 2 = 0$

$x + 3 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

Solve for x square root of x^2+x-2=2

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