Convert to Interval Notation x^2+4x+3>=0

$x^{2} + 4 x + 3 \geq 0$

Step 1

Convert the inequality to an equation.

$x^{2} + 4 x + 3 = 0$

Step 2

Factor $x^{2} + 4 x + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

$(x + 1) (x + 3) = 0$

Step 3

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

$x + 1 = 0$

$x + 3 = 0$

Step 4

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.

$x = - 1$

Step 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.

$x = - 3$

Step 6

The final solution is all the values that make $(x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

Step 7

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$x > - 1$

Step 8

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.

${(- 6)}^{2} + 4 (- 6) + 3 \geq 0$

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = - 2$

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

${(- 2)}^{2} + 4 (- 2) + 3 \geq 0$

The left side $- 1$ 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.

$x = 0$

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

${(0)}^{2} + 4 (0) + 3 \geq 0$

The left side $3$ 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.

$x < - 3$ True

$- 3 < x < - 1$ False

$x > - 1$ True

$x < - 3$ True

$- 3 < x < - 1$ False

$x > - 1$ True

Step 9

The solution consists of all of the true intervals.

$x \leq - 3$ or $x \geq - 1$

Step 10

Convert the inequality to interval notation.

$(- \infty, - 3] \cup [- 1, \infty)$

Step 11

Convert to Interval Notation x^2+4x+3>=0

Related Questions

Support LCMGCF.com by disabling your adblocker.