Grade mathematics: Quadratic Inequalities | PPT - Free Printable

To solve and graph the given inequalities, we will solve each inequality step by step. Here are the solutions:

---

1. \( y - 27y + 70 < 0 \)

Simplify the expression:
\[
y - 27y + 70 = -26y + 70
\]
So the inequality becomes:
\[
-26y + 70 < 0
\]

Solve for \( y \):
\[
-26y < -70
\]
Divide both sides by \(-26\) (and reverse the inequality sign):
\[
y > \frac{70}{26}
\]
Simplify the fraction:
\[
y > \frac{35}{13}
\]

Solution: \( y > \frac{35}{13} \)

---

2. \( x^2 + 2x + 13 > -7 \)

Simplify the inequality:
\[
x^2 + 2x + 13 + 7 > 0
\]
\[
x^2 + 2x + 20 > 0
\]

The quadratic \( x^2 + 2x + 20 \) is always positive because its discriminant is negative:
\[
\Delta = b^2 - 4ac = 2^2 - 4(1)(20) = 4 - 80 = -76
\]
Since the discriminant is negative, the quadratic has no real roots and opens upwards (coefficient of \( x^2 \) is positive). Therefore, it is always positive.

Solution: \( x \in \mathbb{R} \) (all real numbers)

---

3. \( x(x + 1) > 12 - 5x \)

Expand and simplify:
\[
x^2 + x > 12 - 5x
\]
\[
x^2 + x + 5x - 12 > 0
\]
\[
x^2 + 6x - 12 > 0
\]

Solve the quadratic equation \( x^2 + 6x - 12 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4(1)(-12)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 48}}{2} = \frac{-6 \pm \sqrt{84}}{2} = \frac{-6 \pm 2\sqrt{21}}{2} = -3 \pm \sqrt{21}
\]

The roots are:
\[
x = -3 + \sqrt{21} \quad \text{and} \quad x = -3 - \sqrt{21}
\]

The quadratic \( x^2 + 6x - 12 \) opens upwards (coefficient of \( x^2 \) is positive), so the solution to the inequality is:
\[
x < -3 - \sqrt{21} \quad \text{or} \quad x > -3 + \sqrt{21}
\]

Solution: \( x \in (-\infty, -3 - \sqrt{21}) \cup (-3 + \sqrt{21}, \infty) \)

---

4. \( a^2 + 3a + 2 < -3(a + 2) \)

Expand and simplify:
\[
a^2 + 3a + 2 < -3a - 6
\]
\[
a^2 + 3a + 3a + 2 + 6 < 0
\]
\[
a^2 + 6a + 8 < 0
\]

Factor the quadratic:
\[
a^2 + 6a + 8 = (a + 2)(a + 4)
\]
So the inequality becomes:
\[
(a + 2)(a + 4) < 0
\]

Find the critical points:
\[
a = -2 \quad \text{and} \quad a = -4
\]

Test intervals: \( (-\infty, -4) \), \( (-4, -2) \), and \( (-2, \infty) \).

- For \( a \in (-\infty, -4) \): Choose \( a = -5 \):
\[
(-5 + 2)(-5 + 4) = (-3)(-1) = 3 > 0
\]
- For \( a \in (-4, -2) \): Choose \( a = -3 \):
\[
(-3 + 2)(-3 + 4) = (-1)(1) = -1 < 0
\]
- For \( a \in (-2, \infty) \): Choose \( a = 0 \):
\[
(0 + 2)(0 + 4) = (2)(4) = 8 > 0
\]

The solution is:
\[
a \in (-4, -2)
\]

Solution: \( a \in (-4, -2) \)

---

5. \( 2x + 5x = 2 \)

Simplify:
\[
7x = 2
\]
\[
x = \frac{2}{7}
\]

This is an equation, not an inequality. The solution is:
\[
x = \frac{2}{7}
\]

Solution: \( x = \frac{2}{7} \)

---

6. \( 10 - 9y \geq -2y \)

Simplify:
\[
10 - 9y + 2y \geq 0
\]
\[
10 - 7y \geq 0
\]
\[
-7y \geq -10
\]
Divide by \(-7\) (and reverse the inequality sign):
\[
y \leq \frac{10}{7}
\]

Solution: \( y \leq \frac{10}{7} \)

---

7. \( b(b + 3) \geq -2 \)

Expand and simplify:
\[
b^2 + 3b \geq -2
\]
\[
b^2 + 3b + 2 \geq 0
\]

Factor the quadratic:
\[
b^2 + 3b + 2 = (b + 1)(b + 2)
\]
So the inequality becomes:
\[
(b + 1)(b + 2) \geq 0
\]

Find the critical points:
\[
b = -1 \quad \text{and} \quad b = -2
\]

Test intervals: \( (-\infty, -2) \), \( (-2, -1) \), and \( (-1, \infty) \).

- For \( b \in (-\infty, -2) \): Choose \( b = -3 \):
\[
(-3 + 1)(-3 + 2) = (-2)(-1) = 2 > 0
\]
- For \( b \in (-2, -1) \): Choose \( b = -1.5 \):
\[
(-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) = -0.25 < 0
\]
- For \( b \in (-1, \infty) \): Choose \( b = 0 \):
\[
(0 + 1)(0 + 2) = (1)(2) = 2 > 0
\]

The solution is:
\[
b \in (-\infty, -2] \cup [-1, \infty)
\]

Solution: \( b \in (-\infty, -2] \cup [-1, \infty) \)

---

8. \( a \geq 4(2a - 3) \)

Expand and simplify:
\[
a \geq 8a - 12
\]
\[
a - 8a \geq -12
\]
\[
-7a \geq -12
\]
Divide by \(-7\) (and reverse the inequality sign):
\[
a \leq \frac{12}{7}
\]

Solution: \( a \leq \frac{12}{7} \)

---

9. \( y - 27y + 70 > 0 \)

This is the same as Problem 1, but with the inequality reversed:
\[
-26y + 70 > 0
\]
\[
-26y > -70
\]
\[
y < \frac{70}{26}
\]
\[
y < \frac{35}{13}
\]

Solution: \( y < \frac{35}{13} \)

---

10. \( x^2 + 9x + 13 > -7 \)

This is the same as Problem 2:
\[
x^2 + 9x + 20 > 0
\]

The quadratic \( x^2 + 9x + 20 \) is always positive because its discriminant is negative:
\[
\Delta = 9^2 - 4(1)(20) = 81 - 80 = 1
\]
Since the discriminant is positive, the quadratic has two real roots, but it opens upwards (coefficient of \( x^2 \) is positive). Therefore, it is always positive.

Solution: \( x \in \mathbb{R} \) (all real numbers)

---

Final Answers:

1. \( y > \frac{35}{13} \)
2. \( x \in \mathbb{R} \)
3. \( x \in (-\infty, -3 - \sqrt{21}) \cup (-3 + \sqrt{21}, \infty) \)
4. \( a \in (-4, -2) \)
5. \( x = \frac{2}{7} \)
6. \( y \leq \frac{10}{7} \)
7. \( b \in (-\infty, -2] \cup [-1, \infty) \)
8. \( a \leq \frac{12}{7} \)
9. \( y < \frac{35}{13} \)
10. \( x \in \mathbb{R} \)

\[
\boxed{
\begin{aligned}
1. & \ y > \frac{35}{13} \\
2. & \ x \in \mathbb{R} \\
3. & \ x \in (-\infty, -3 - \sqrt{21}) \cup (-3 + \sqrt{21}, \infty) \\
4. & \ a \in (-4, -2) \\
5. & \ x = \frac{2}{7} \\
6. & \ y \leq \frac{10}{7} \\
7. & \ b \in (-\infty, -2] \cup [-1, \infty) \\
8. & \ a \leq \frac{12}{7} \\
9. & \ y < \frac{35}{13} \\
10. & \ x \in \mathbb{R}
\end{aligned}
}
\]