Solve the Inequality for x square root of x^2+4x-5<4

$\sqrt{x^{2} + 4 x - 5} < 4$

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

${\sqrt{x^{2} + 4 x - 5}}^{2} < 4^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + 4 x - 5}$ as ${(x^{2} + 4 x - 5)}^{\frac{1}{2}}$ .

${({(x^{2} + 4 x - 5)}^{\frac{1}{2}})}^{2} < 4^{2}$

Simplify the left side.

Simplify ${({(x^{2} + 4 x - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x^{2} + 4 x - 5)}^{\frac{1}{2}})}^{2}$ .

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

${(x^{2} + 4 x - 5)}^{\frac{1}{2} \cdot 2} < 4^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x^{2} + 4 x - 5)}^{\frac{1}{2} \cdot 2} < 4^{2}$

Rewrite the expression.

${(x^{2} + 4 x - 5)}^{1} < 4^{2}$

Simplify.

$x^{2} + 4 x - 5 < 4^{2}$

Simplify the right side.

Raise $4$ to the power of $2$ .

$x^{2} + 4 x - 5 < 16$

Solve for $x$ .

Convert the inequality to an equation.

$x^{2} + 4 x - 5 = 16$

Subtract $16$ from both sides of the equation.

$x^{2} + 4 x - 5 - 16 = 0$

Subtract $16$ from $- 5$ .

$x^{2} + 4 x - 21 = 0$

Factor $x^{2} + 4 x - 21$ 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 $- 21$ and whose sum is $4$ .

$- 3, 7$

Write the factored form using these integers.

$(x - 3) (x + 7) = 0$

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

$x - 3 = 0$

$x + 7 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x - 3) (x + 7) = 0$ true.

$x = 3, - 7$

Find the domain of $\sqrt{(x - 1) (x + 5)} - 4$ .

Set the radicand in $\sqrt{(x - 1) (x + 5)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 1) (x + 5) \geq 0$

Solve for $x$ .

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

$x - 1 = 0$

$x + 5 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 1) (x + 5) \geq 0$ true.

$x = 1, - 5$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$x > 1$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$((- 8) - 1) ((- 8) + 5) \geq 0$

The left side $27$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$((0) - 1) ((0) + 5) \geq 0$

The left side $- 5$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$((4) - 1) ((4) + 5) \geq 0$

The left side $27$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 5$ True

$- 5 < x < 1$ False

$x > 1$ True

$x < - 5$ True

$- 5 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$x \leq - 5$ or $x \geq 1$

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

$(- \infty, - 5] \cup [1, \infty)$

Use each root to create test intervals.

$x < - 7$

$- 7 < x < - 5$

$- 5 < x < 1$

$1 < x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$\sqrt{{(- 10)}^{2} + 4 (- 10) - 5} < 4$

The left side $7.41619848$ is not less than the right side $4$ , which means that the given statement is false.

False

Test a value on the interval $- 7 < x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < - 5$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$\sqrt{{(- 6)}^{2} + 4 (- 6) - 5} < 4$

The left side $2.64575131$ is less than the right side $4$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\sqrt{{(0)}^{2} + 4 (0) - 5} < 4$

The left side is not equal to the right side, which means that the given statement is false.

False

Test a value on the interval $1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$\sqrt{{(2)}^{2} + 4 (2) - 5} < 4$

The left side $2.64575131$ is less than the right side $4$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\sqrt{{(6)}^{2} + 4 (6) - 5} < 4$

The left side $7.41619848$ is not less than the right side $4$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 7$ False

$- 7 < x < - 5$ True

$- 5 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

$x < - 7$ False

$- 7 < x < - 5$ True

$- 5 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 7 < x < - 5$ or $1 < x < 3$

The result can be shown in multiple forms.

Inequality Form:

$- 7 < x < - 5 or 1 < x < 3$

Interval Notation:

$(- 7, - 5) \cup (1, 3)$

Solve the Inequality for x square root of x^2+4x-5<4

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