Solve the Inequality for x square root of x-3>5-x

$\sqrt{x - 3} > 5 - x$

Step 1

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

${\sqrt{x - 3}}^{2} > {(5 - x)}^{2}$

Step 2

Simplify each side of the inequality.

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

${({(x - 3)}^{\frac{1}{2}})}^{2} > {(5 - x)}^{2}$

Simplify the left side.

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

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

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

${(x - 3)}^{\frac{1}{2} \cdot 2} > {(5 - x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x - 3)}^{\frac{1}{2} \cdot 2} > {(5 - x)}^{2}$

Rewrite the expression.

${(x - 3)}^{1} > {(5 - x)}^{2}$

Simplify.

$x - 3 > {(5 - x)}^{2}$

Simplify the right side.

Simplify ${(5 - x)}^{2}$ .

Rewrite ${(5 - x)}^{2}$ as $(5 - x) (5 - x)$ .

$x - 3 > (5 - x) (5 - x)$

Expand $(5 - x) (5 - x)$ using the FOIL Method.

Apply the distributive property.

$x - 3 > 5 (5 - x) - x (5 - x)$

Apply the distributive property.

$x - 3 > 5 \cdot 5 + 5 (- x) - x (5 - x)$

Apply the distributive property.

$x - 3 > 5 \cdot 5 + 5 (- x) - x \cdot 5 - x (- x)$

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$x - 3 > 25 + 5 (- x) - x \cdot 5 - x (- x)$

Multiply $- 1$ by $5$ .

$x - 3 > 25 - 5 x - x \cdot 5 - x (- x)$

Multiply $5$ by $- 1$ .

$x - 3 > 25 - 5 x - 5 x - x (- x)$

Rewrite using the commutative property of multiplication.

$x - 3 > 25 - 5 x - 5 x - 1 \cdot - 1 x \cdot x$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x - 3 > 25 - 5 x - 5 x - 1 \cdot - 1 (x \cdot x)$

Multiply $x$ by $x$ .

$x - 3 > 25 - 5 x - 5 x - 1 \cdot - 1 x^{2}$

Multiply $- 1$ by $- 1$ .

$x - 3 > 25 - 5 x - 5 x + 1 x^{2}$

Multiply $x^{2}$ by $1$ .

$x - 3 > 25 - 5 x - 5 x + x^{2}$

Subtract $5 x$ from $- 5 x$ .

$x - 3 > 25 - 10 x + x^{2}$

Step 3

Solve for $x$ .

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

$25 - 10 x + x^{2} < x - 3$

Move all terms containing $x$ to the left side of the inequality.

Subtract $x$ from both sides of the inequality.

$25 - 10 x + x^{2} - x < - 3$

Subtract $x$ from $- 10 x$ .

$25 + x^{2} - 11 x < - 3$

Convert the inequality to an equation.

$25 + x^{2} - 11 x = - 3$

Add $3$ to both sides of the equation.

$25 + x^{2} - 11 x + 3 = 0$

Add $25$ and $3$ .

$x^{2} - 11 x + 28 = 0$

Factor $x^{2} - 11 x + 28$ 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 $28$ and whose sum is $- 11$ .

$- 7, - 4$

Write the factored form using these integers.

$(x - 7) (x - 4) = 0$

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

$x - 7 = 0$

$x - 4 = 0$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(x - 7) (x - 4) = 0$ true.

$x = 7, 4$

Step 4

Find the domain of $\sqrt{x - 3} - 5 + x$ .

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

$x - 3 \geq 0$

Add $3$ to both sides of the inequality.

$x \geq 3$

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

$[3, \infty)$

Step 5

Use each root to create test intervals.

$x < 3$

$3 < x < 4$

$4 < x < 7$

$x > 7$

Step 6

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 $x < 3$ 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.

$x = 0$

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

$\sqrt{(0) - 3} > 5 - (0)$

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 $3 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 4$ and see if this value makes the original inequality true.

$x = 3.5$

Replace $x$ with $3.5$ in the original inequality.

$\sqrt{(3.5) - 3} > 5 - (3.5)$

The left side $0.70710678$ is not greater than the right side $1.5$ , which means that the given statement is false.

False

Test a value on the interval $4 < x < 7$ to see if it makes the inequality true.

Choose a value on the interval $4 < x < 7$ and see if this value makes the original inequality true.

$x = 6$

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

$\sqrt{(6) - 3} > 5 - (6)$

The left side $1.7320508$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

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.

$\sqrt{(10) - 3} > 5 - (10)$

The left side $2.64575131$ is greater than the right side $- 5$ , which means that the given statement is always true.

True

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

$x < 3$ False

$3 < x < 4$ False

$4 < x < 7$ True

$x > 7$ True

$x < 3$ False

$3 < x < 4$ False

$4 < x < 7$ True

$x > 7$ True

Step 7

The solution consists of all of the true intervals.

$4 < x < 7$ or $x > 7$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$4 < x < 7 or x > 7$

Interval Notation:

$(4, 7) \cup (7, \infty)$

Step 9

Solve the Inequality for x square root of x-3>5-x

Related Questions

Support LCMGCF.com by disabling your adblocker.