Solve the Inequality for x square root of x^2-3x+2< = square root of x+7

$\sqrt{x^{2} - 3 x + 2} \leq \sqrt{x + 7}$

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

${\sqrt{x^{2} - 3 x + 2}}^{2} \leq {\sqrt{x + 7}}^{2}$

Simplify each side of the inequality.

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

${({(x^{2} - 3 x + 2)}^{\frac{1}{2}})}^{2} \leq {\sqrt{x + 7}}^{2}$

Simplify the left side.

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

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

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

${(x^{2} - 3 x + 2)}^{\frac{1}{2} \cdot 2} \leq {\sqrt{x + 7}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x^{2} - 3 x + 2)}^{\frac{1}{2} \cdot 2} \leq {\sqrt{x + 7}}^{2}$

Rewrite the expression.

${(x^{2} - 3 x + 2)}^{1} \leq {\sqrt{x + 7}}^{2}$

Simplify.

$x^{2} - 3 x + 2 \leq {\sqrt{x + 7}}^{2}$

Simplify the right side.

Rewrite ${\sqrt{x + 7}}^{2}$ as $x + 7$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 7}$ as ${(x + 7)}^{\frac{1}{2}}$ .

$x^{2} - 3 x + 2 \leq {({(x + 7)}^{\frac{1}{2}})}^{2}$

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

$x^{2} - 3 x + 2 \leq {(x + 7)}^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$x^{2} - 3 x + 2 \leq {(x + 7)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{2} - 3 x + 2 \leq {(x + 7)}^{\frac{2}{2}}$

Rewrite the expression.

$x^{2} - 3 x + 2 \leq {(x + 7)}^{1}$

Simplify.

$x^{2} - 3 x + 2 \leq x + 7$

Solve for $x$ .

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

Subtract $x$ from both sides of the inequality.

$x^{2} - 3 x + 2 - x \leq 7$

Subtract $x$ from $- 3 x$ .

$x^{2} - 4 x + 2 \leq 7$

Convert the inequality to an equation.

$x^{2} - 4 x + 2 = 7$

Subtract $7$ from both sides of the equation.

$x^{2} - 4 x + 2 - 7 = 0$

Subtract $7$ from $2$ .

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

Factor $x^{2} - 4 x - 5$ 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 $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Write the factored form using these integers.

$(x - 5) (x + 1) = 0$

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

$x - 5 = 0$

$x + 1 = 0$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = 5, - 1$

Find the domain of $\sqrt{(x - 2) (x - 1)} - \sqrt{x + 7}$ .

Set the radicand in $\sqrt{(x - 2) (x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 2) (x - 1) \geq 0$

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 $0$ .

$x - 2 = 0$

$x - 1 = 0$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 2) (x - 1) \geq 0$ true.

$x = 2, 1$

Use each root to create test intervals.

$x < 1$

$1 < x < 2$

$x > 2$

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 $x < 1$ 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.

$x = 0$

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

$((0) - 2) ((0) - 1) \geq 0$

The left side $2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Replace $x$ with $1.5$ in the original inequality.

$((1.5) - 2) ((1.5) - 1) \geq 0$

The left side $- 0.25$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

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

$((4) - 2) ((4) - 1) \geq 0$

The left side $6$ 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.

$x < 1$ True

$1 < x < 2$ False

$x > 2$ True

$x < 1$ True

$1 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$x \leq 1$ or $x \geq 2$

Set the radicand in $\sqrt{x + 7}$ greater than or equal to $0$ to find where the expression is defined.

$x + 7 \geq 0$

Subtract $7$ from both sides of the inequality.

$x \geq - 7$

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

$[- 7, 1] \cup [2, \infty)$

Use each root to create test intervals.

$x < - 7$

$- 7 < x < - 1$

$- 1 < x < 1$

$1 < x < 2$

$2 < x < 5$

$x > 5$

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.

$\sqrt{{(- 10)}^{2} - 3 \cdot - 10 + 2} \leq \sqrt{(- 10) + 7}$

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 $- 7 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

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

$\sqrt{{(- 4)}^{2} - 3 \cdot - 4 + 2} \leq \sqrt{(- 4) + 7}$

The left side $5.47722557$ is greater than the right side $1.7320508$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$\sqrt{{(0)}^{2} - 3 \cdot 0 + 2} \leq \sqrt{(0) + 7}$

The left side $1.41421356$ is less than the right side $2.64575131$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Replace $x$ with $1.5$ in the original inequality.

$\sqrt{{(1.5)}^{2} - 3 \cdot 1.5 + 2} \leq \sqrt{(1.5) + 7}$

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 $2 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

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

$\sqrt{{(4)}^{2} - 3 \cdot 4 + 2} \leq \sqrt{(4) + 7}$

The left side $2.44948974$ is less than the right side $3.31662479$ , which means that the given statement is always true.

True

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.

$\sqrt{{(8)}^{2} - 3 \cdot 8 + 2} \leq \sqrt{(8) + 7}$

The left side $6.48074069$ is greater than the right side $3.87298334$ , which means that the given statement is false.

False

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

$x < - 7$ False

$- 7 < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 2$ False

$2 < x < 5$ True

$x > 5$ False

$x < - 7$ False

$- 7 < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 2$ False

$2 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$- 1 \leq x \leq 1$ or $2 \leq x \leq 5$

The result can be shown in multiple forms.

Inequality Form:

$- 1 \leq x \leq 1 or 2 \leq x \leq 5$

Interval Notation:

$[- 1, 1] \cup [2, 5]$

Solve the Inequality for x square root of x^2-3x+2< = square root of x+7

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