Solve the Inequality for x 100x^2-20x+1<0

$100 x^{2} - 20 x + 1 < 0$

Step 1

Convert the inequality to an equation.

$100 x^{2} - 20 x + 1 = 0$

Step 2

Factor using the perfect square rule.

Rewrite $100 x^{2}$ as ${(10 x)}^{2}$ .

${(10 x)}^{2} - 20 x + 1 = 0$

Rewrite $1$ as $1^{2}$ .

${(10 x)}^{2} - 20 x + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$20 x = 2 \cdot (10 x) \cdot 1$

Rewrite the polynomial.

${(10 x)}^{2} - 2 \cdot (10 x) \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 10 x$ and $b = 1$ .

${(10 x - 1)}^{2} = 0$

Step 3

Set the $10 x - 1$ equal to $0$ .

$10 x - 1 = 0$

Step 4

Solve for $x$ .

Add $1$ to both sides of the equation.

$10 x = 1$

Divide each term in $10 x = 1$ by $10$ and simplify.

Divide each term in $10 x = 1$ by $10$ .

$\frac{10 x}{10} = \frac{1}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 x}{10} = \frac{1}{10}$

Divide $x$ by $1$ .

$x = \frac{1}{10}$

Step 5

Use each root to create test intervals.

$x < \frac{1}{10}$

$x > \frac{1}{10}$

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 $x < \frac{1}{10}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{1}{10}$ and see if this value makes the original inequality true.

$x = 0$

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

$100 {(0)}^{2} - 20 \cdot 0 + 1 < 0$

The left side $1$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{1}{10}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{1}{10}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$100 {(3)}^{2} - 20 \cdot 3 + 1 < 0$

The left side $841$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$x < \frac{1}{10}$ False

$x > \frac{1}{10}$ False

$x < \frac{1}{10}$ False

$x > \frac{1}{10}$ False

Step 7

Since there are no numbers that fall within the interval, this inequality has no solution.

No solution

Solve the Inequality for x 100x^2-20x+1<0

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