Inequality Solver worksheet

1. Solution (interval notation): (−∞, 5 − 2√5) ∪ (5 + 2√5, ∞)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{x² − 10x + 5 > 0}$

2. Find the discriminant, D = b² − 4ac (a, b and c are the numbers in front of x², x and on their own). It tells how many times the graph of the left side crosses zero

   $D$$={\left(\mathrm{-10}\right)}^2-4\times 1\times 5$$=80$

3. D is above 0, so the graph crosses zero twice. Find those two crossing points (roots) with the quadratic formula

   $\text{x = 5 − 2√5 or x = 5 + 2√5 (≈ 0.527864 and 9.472136)}$

4. The roots split the number line into three pieces. Test a point in each. The x² term is positive, so the left side is positive outside the roots and negative between them

   $\text{x = −1: 16 (+); x = 5: −20 (−); x = 11: 16 (+)}$

5. Keep the pieces where the left side is positive

   $\text{x < 5 − 2√5 or x > 5 + 2√5: (−∞, 5 − 2√5) ∪ (5 + 2√5, ∞)}$

2. Solution (interval notation): (−∞, 0.5)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{−4x + 2 > 0}$

2. Move the plain number to the right side, leaving the x term on its own

   $\text{−4x > −2}$

3. Divide both sides by −4. Dividing by a negative number reverses the inequality sign. The boundary value is

   $\frac{\mathrm{-2}}{\mathrm{-4}}$$=0.5$

4. So x < 0.5. In interval notation, a round bracket leaves the endpoint out and a square bracket includes it

   $\text{x < 0.5, so (−∞, 0.5)}$

3. Solution (interval notation): (−∞, ∞)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{x² − x + 3 > 0}$

2. Find the discriminant, D = b² − 4ac (a, b and c are the numbers in front of x², x and on their own). It tells how many times the graph of the left side crosses zero

   $D$$={\left(\mathrm{-1}\right)}^2-4\times 1\times 3$$=\mathrm{-11}$

3. D is below 0, so the graph never reaches 0. With a positive x² term, the left side is positive for every x

   $\text{x² − x + 3 > 0 for all x}$

4. Keep the pieces where the left side is positive

   $\text{All real numbers: (−∞, ∞)}$

4. Solution (interval notation): (−∞, 0) ∪ (2, ∞)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{x² − 2x > 0}$

2. Find the discriminant, D = b² − 4ac (a, b and c are the numbers in front of x², x and on their own). It tells how many times the graph of the left side crosses zero

   $D$$={\left(\mathrm{-2}\right)}^2-4\times 1\times 0$$=4$

3. D is above 0, so the graph crosses zero twice. Find those two crossing points (roots) with the quadratic formula

   $\text{x = 0 or x = 2 (≈ 0 and 2)}$

4. The roots split the number line into three pieces. Test a point in each. The x² term is positive, so the left side is positive outside the roots and negative between them

   $\text{x = −1: 3 (+); x = 1: −1 (−); x = 3: 3 (+)}$

5. Keep the pieces where the left side is positive

   $\text{x < 0 or x > 2: (−∞, 0) ∪ (2, ∞)}$

5. Solution (interval notation): (−∞, 1/7)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{−7x + 1 > 0}$

2. Move the plain number to the right side, leaving the x term on its own

   $\text{−7x > −1}$

3. Divide both sides by −7. Dividing by a negative number reverses the inequality sign. The boundary value is

   $\frac{\mathrm{-1}}{\mathrm{-7}}$$=0.142857$

4. So x < 1/7. In interval notation, a round bracket leaves the endpoint out and a square bracket includes it

   $\text{x < 1/7, so (−∞, 1/7)}$

6. Solution (interval notation): (−∞, 0) ∪ (3, ∞)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{x² − 3x > 0}$

2. Find the discriminant, D = b² − 4ac (a, b and c are the numbers in front of x², x and on their own). It tells how many times the graph of the left side crosses zero

   $D$$={\left(\mathrm{-3}\right)}^2-4\times 1\times 0$$=9$

3. D is above 0, so the graph crosses zero twice. Find those two crossing points (roots) with the quadratic formula

   $\text{x = 0 or x = 3 (≈ 0 and 3)}$

4. The roots split the number line into three pieces. Test a point in each. The x² term is positive, so the left side is positive outside the roots and negative between them

   $\text{x = −1: 4 (+); x = 1.5: −2.25 (−); x = 4: 4 (+)}$

5. Keep the pieces where the left side is positive

   $\text{x < 0 or x > 3: (−∞, 0) ∪ (3, ∞)}$

7. Solution (interval notation): (−∞, ∞)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{x² − 2x + 4 > 0}$

2. Find the discriminant, D = b² − 4ac (a, b and c are the numbers in front of x², x and on their own). It tells how many times the graph of the left side crosses zero

   $D$$={\left(\mathrm{-2}\right)}^2-4\times 1\times 4$$=\mathrm{-12}$

3. D is below 0, so the graph never reaches 0. With a positive x² term, the left side is positive for every x

   $\text{x² − 2x + 4 > 0 for all x}$

4. Keep the pieces where the left side is positive

   $\text{All real numbers: (−∞, ∞)}$

8. Solution (interval notation): (−∞, (9 − √69)/2) ∪ ((9 + √69)/2, ∞)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{x² − 9x + 3 > 0}$

2. Find the discriminant, D = b² − 4ac (a, b and c are the numbers in front of x², x and on their own). It tells how many times the graph of the left side crosses zero

   $D$$={\left(\mathrm{-9}\right)}^2-4\times 1\times 3$$=69$

3. D is above 0, so the graph crosses zero twice. Find those two crossing points (roots) with the quadratic formula

   $\text{x = (9 − √69)/2 or x = (9 + √69)/2 (≈ 0.346688 and 8.653312)}$

4. The roots split the number line into three pieces. Test a point in each. The x² term is positive, so the left side is positive outside the roots and negative between them

   $\text{x = −1: 13 (+); x = 4.5: −17.25 (−); x = 10: 13 (+)}$

5. Keep the pieces where the left side is positive

   $\text{x < (9 − √69)/2 or x > (9 + √69)/2: (−∞, (9 − √69)/2) ∪ ((9 + √69)/2, ∞)}$

9. Solution (interval notation): (−∞, ∞)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{2 > 0}$

2. The x terms cancel, leaving 2 > 0, which is always true, so every x works

   $\text{All real numbers: (−∞, ∞)}$

10. Solution (interval notation): (−∞, 2/7)

1. Move every term to the left side, so we only need to ask when the left side is above or below 0

   $\text{−7x + 2 > 0}$

2. Move the plain number to the right side, leaving the x term on its own

   $\text{−7x > −2}$

3. Divide both sides by −7. Dividing by a negative number reverses the inequality sign. The boundary value is

   $\frac{\mathrm{-2}}{\mathrm{-7}}$$=0.285714$

4. So x < 2/7. In interval notation, a round bracket leaves the endpoint out and a square bracket includes it

   $\text{x < 2/7, so (−∞, 2/7)}$