Solve for x square root of x^2-2x+1>=4

$\sqrt{x^{2} - 2 x + 1} \geq 4$

Step 1

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

${\sqrt{x^{2} - 2 x + 1}}^{2} \geq 4^{2}$

Step 2

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

${(x^{2} - 2 x + 1)}^{1} \geq 4^{2}$

Simplify.

$x^{2} - 2 x + 1 \geq 4^{2}$

Simplify the right side.

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

$x^{2} - 2 x + 1 \geq 16$

Step 3

Solve for $x$ .

Convert the inequality to an equation.

$x^{2} - 2 x + 1 = 16$

Subtract $16$ from both sides of the equation.

$x^{2} - 2 x + 1 - 16 = 0$

Subtract $16$ from $1$ .

$x^{2} - 2 x - 15 = 0$

Factor $x^{2} - 2 x - 15$ 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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$(x - 5) (x + 3) = 0$

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

$x - 5 = 0$

$x + 3 = 0$

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 $x + 3$ 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 $(x - 5) (x + 3) = 0$ true.

$x = 5, - 3$

Step 4

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 5$

$x > 5$

Step 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 $x < - 3$ 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.

$\sqrt{{(- 6)}^{2} - 2 \cdot - 6 + 1} \geq 4$

The left side $7$ is greater than the right side $4$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

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

$\sqrt{{(0)}^{2} - 2 \cdot 0 + 1} \geq 4$

The left side $1$ is less than the right side $4$ , which means that the given statement is false.

False

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.

$\sqrt{{(8)}^{2} - 2 \cdot 8 + 1} \geq 4$

The left side $7$ is greater than the right side $4$ , which means that the given statement is always true.

True

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

$x < - 3$ True

$- 3 < x < 5$ False

$x > 5$ True

$x < - 3$ True

$- 3 < x < 5$ False

$x > 5$ True

Step 6

The solution consists of all of the true intervals.

$x \leq - 3$ or $x \geq 5$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or x \geq 5$

Interval Notation:

$(- \infty, - 3] \cup [5, \infty)$

Step 8

Solve for x square root of x^2-2x+1>=4

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