Solve the Inequality for x x^2+3>y

$x^{2} + 3 > y$

Subtract $3$ from both sides of the inequality.

$x^{2} > y - 3$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} > \sqrt{y - 3}$

Simplify the left side.

Pull terms out from under the radical.

$| x | > \sqrt{y - 3}$

Write $| x | > \sqrt{y - 3}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x > \sqrt{y - 3}$

Find the domain of $x > \sqrt{y - 3}$ and find the intersection with $x \geq 0$ .

Find the domain of $x > \sqrt{y - 3}$ .

Set the radicand in $\sqrt{y - 3}$ greater than or equal to $0$ to find where the expression is defined.

$y - 3 \geq 0$

Add $3$ to both sides of the inequality.

$y \geq 3$

The domain is all values of $x$ that make the expression defined.

$[3, \infty)$

Find the intersection of $x \geq 0$ and $[3, \infty)$ .

$x \geq 3$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x > \sqrt{y - 3}$

Find the domain of $- x > \sqrt{y - 3}$ and find the intersection with $x < 0$ .

Find the domain of $- x > \sqrt{y - 3}$ .

Set the radicand in $\sqrt{y - 3}$ greater than or equal to $0$ to find where the expression is defined.

$y - 3 \geq 0$

Add $3$ to both sides of the inequality.

$y \geq 3$

The domain is all values of $x$ that make the expression defined.

$[3, \infty)$

Find the intersection of $x < 0$ and $[3, \infty)$ .

$No solution$

Write as a piecewise.

${\begin{cases} x > \sqrt{y - 3} & x \geq 3 \end{cases}$

Solve $x > \sqrt{y - 3}$ when $x \geq 3$ .

Solve $x > \sqrt{y - 3}$ for $y$ .

Rewrite so $y$ is on the left side of the inequality.

$\sqrt{y - 3} < x$

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

${\sqrt{y - 3}}^{2} < x^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y - 3}$ as ${(y - 3)}^{\frac{1}{2}}$ .

${({(y - 3)}^{\frac{1}{2}})}^{2} < x^{2}$

Simplify the left side.

Simplify ${({(y - 3)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(y - 3)}^{\frac{1}{2}})}^{2}$ .

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

${(y - 3)}^{\frac{1}{2} \cdot 2} < x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y - 3)}^{\frac{1}{2} \cdot 2} < x^{2}$

Rewrite the expression.

${(y - 3)}^{1} < x^{2}$

Simplify.

$y - 3 < x^{2}$

Add $3$ to both sides of the inequality.

$y < x^{2} + 3$

Find the intersection of $y < x^{2} + 3$ and $x \geq 3$ .

$3 \leq x < x^{2} + 3$

Find the union of the solutions.

$3 \leq x < x^{2} + 3$

The result can be shown in multiple forms.

Inequality Form:

$3 \leq x < x^{2} + 3$

Interval Notation:

$[3, x^{2} + 3)$

Solve the Inequality for x x^2+3>y

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