Solve the Inequality for x (x^2+4x-3)/(x^2+1)<x

$\frac{x^{2} + 4 x - 3}{x^{2} + 1} < x$

Step 1

Subtract $x$ from both sides of the inequality.

$\frac{x^{2} + 4 x - 3}{x^{2} + 1} - x < 0$

Step 2

Simplify $\frac{x^{2} + 4 x - 3}{x^{2} + 1} - x$ .

To write $- x$ as a fraction with a common denominator, multiply by $\frac{x^{2} + 1}{x^{2} + 1}$ .

$\frac{x^{2} + 4 x - 3}{x^{2} + 1} - x \cdot \frac{x^{2} + 1}{x^{2} + 1} < 0$

Combine $- x$ and $\frac{x^{2} + 1}{x^{2} + 1}$ .

$\frac{x^{2} + 4 x - 3}{x^{2} + 1} + \frac{- x (x^{2} + 1)}{x^{2} + 1} < 0$

Combine the numerators over the common denominator.

$\frac{x^{2} + 4 x - 3 - x (x^{2} + 1)}{x^{2} + 1} < 0$

Simplify the numerator.

Apply the distributive property.

$\frac{x^{2} + 4 x - 3 - x \cdot x^{2} - x \cdot 1}{x^{2} + 1} < 0$

Multiply $x$ by $x^{2}$ by adding the exponents.

Move $x^{2}$ .

$\frac{x^{2} + 4 x - 3 - (x^{2} x) - x \cdot 1}{x^{2} + 1} < 0$

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

Raise $x$ to the power of $1$ .

$\frac{x^{2} + 4 x - 3 - (x^{2} x^{1}) - x \cdot 1}{x^{2} + 1} < 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{x^{2} + 4 x - 3 - x^{2 + 1} - x \cdot 1}{x^{2} + 1} < 0$

Add $2$ and $1$ .

$\frac{x^{2} + 4 x - 3 - x^{3} - x \cdot 1}{x^{2} + 1} < 0$

Multiply $- 1$ by $1$ .

$\frac{x^{2} + 4 x - 3 - x^{3} - x}{x^{2} + 1} < 0$

Subtract $x$ from $4 x$ .

$\frac{x^{2} + 3 x - 3 - x^{3}}{x^{2} + 1} < 0$

Reorder terms.

$\frac{- x^{3} + x^{2} + 3 x - 3}{x^{2} + 1} < 0$

Rewrite $- x^{3} + x^{2} + 3 x - 3$ in a factored form.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(- x^{3} + x^{2}) + 3 x - 3}{x^{2} + 1} < 0$

Factor out the greatest common factor (GCF) from each group.

$\frac{x^{2} (- x + 1) - 3 (- x + 1)}{x^{2} + 1} < 0$

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

$\frac{(- x + 1) (x^{2} - 3)}{x^{2} + 1} < 0$

Factor $- 1$ out of $- x$ .

$\frac{(- (x) + 1) (x^{2} - 3)}{x^{2} + 1} < 0$

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{(- (x) - 1 (- 1)) (x^{2} - 3)}{x^{2} + 1} < 0$

Factor $- 1$ out of $- (x) - 1 (- 1)$ .

$\frac{- (x - 1) (x^{2} - 3)}{x^{2} + 1} < 0$

Rewrite $- (x - 1)$ as $- 1 (x - 1)$ .

$\frac{- 1 (x - 1) (x^{2} - 3)}{x^{2} + 1} < 0$

Move the negative in front of the fraction.

$- \frac{(x - 1) (x^{2} - 3)}{x^{2} + 1} < 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x - 1 = 0$

$x^{2} - 3 = 0$

$x^{2} + 1 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Add $3$ to both sides of the equation.

$x^{2} = 3$

Step 6

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{3}$

Step 7

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{3}, - \sqrt{3}$

Step 8

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 9

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 1}$

Step 10

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 11

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 12

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 1$

$x = \sqrt{3}, - \sqrt{3}$

$x = i, - i$

Step 13

Consolidate the solutions.

$x = 1, \sqrt{3}, - \sqrt{3}$

Step 14

Use each root to create test intervals.

$x < - \sqrt{3}$

$- \sqrt{3} < x < 1$

$1 < x < \sqrt{3}$

$x > \sqrt{3}$

Step 15

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 $x < - \sqrt{3}$ to see if it makes the inequality true.

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

$x = - 4$

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

$\frac{{(- 4)}^{2} + 4 (- 4) - 3}{{(- 4)}^{2} + 1} < - 4$

The left side $- 0.17647058$ is not less than the right side $- 4$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$\frac{{(0)}^{2} + 4 (0) - 3}{{(0)}^{2} + 1} < 0$

The left side $- 3$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 1.37$

Replace $x$ with $1.37$ in the original inequality.

$\frac{{(1.37)}^{2} + 4 (1.37) - 3}{{(1.37)}^{2} + 1} < 1.37$

The left side $1.51444262$ is not less than the right side $1.37$ , which means that the given statement is false.

False

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

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

$x = 4$

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

$\frac{{(4)}^{2} + 4 (4) - 3}{{(4)}^{2} + 1} < 4$

The left side $1.70588235$ is less 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.

$x < - \sqrt{3}$ False

$- \sqrt{3} < x < 1$ True

$1 < x < \sqrt{3}$ False

$x > \sqrt{3}$ True

$x < - \sqrt{3}$ False

$- \sqrt{3} < x < 1$ True

$1 < x < \sqrt{3}$ False

$x > \sqrt{3}$ True

Step 16

The solution consists of all of the true intervals.

$- \sqrt{3} < x < 1$ or $x > \sqrt{3}$

Step 17

The result can be shown in multiple forms.

Inequality Form:

$- \sqrt{3} < x < 1 or x > \sqrt{3}$

Interval Notation:

$(- \sqrt{3}, 1) \cup (\sqrt{3}, \infty)$

Step 18

Solve the Inequality for x (x^2+4x-3)/(x^2+1)<x

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