**Solving Quadratic Inequalities Algebraically : **

In this section, we will learn, how to solve quadratic inequalities algebraically.

**Shortcut 1 : **

**Other Cases :**

If both (ax^{2} + bx + c) and x^{2} have same signs, then there is solution for the given inequality.

If (ax^{2} + bx + c) and x^{2} have different signs,we have to find the value of (b^{2} - 4ac). Then we can know whether there is solution or not by using the hints given on the above tables.

From shortcut 1, if there is solution for the given quadratic inequality, then follow shortcut 2 to know the type of solution.

**Shortcut 2 :**

**Example 1 :**

Solve the quadratic inequality given.

x^{2} + 5x + 6 ≥ 0

**Solution : **

Both (x^{2} + 5x +6) and x^{2} have same signs (positive). So, there is solution for the given inequality.

To know the type of solution, let us find the value of

b^{2} -4ac

b^{2} - 4ac = 5^{2} - 4(1)(6)

b^{2} - 4ac = 25 - 24

b^{2} - 4ac = 1

We get,

b^{2} - 4ac = 1 -----> b^{2} - 4ac ≥ 0

Therefore, there is interval solution.

Let us find the interval solution.

Let x^{2} + 5x + 6 = 0 and solve for x.

x^{2} + 5x + 6 = 0

(x + 2)(x + 3) = 0

x = -2 or x = -3

Let us mark x = -2 and x = -3 on the real number line.

**Testing the interval (-∞, -3] : **

Take a random value in the interval (-∞, -3], say

x = -4

Substitute -4 for x in the given inequality.

x^{2} + 5x + 6 = (-4)^{2} + 5(-4) + 6

x^{2} + 5x + 6 = 16 - 20 + 6

x^{2} + 5x + 6 = 2 **≥ **0

The interval (-∞, -3] satisfies the given inequality.

**Testing the interval [-3, -2] : **

Take a random value in the interval [-3, -2], say

x = -2.5

Substitute -2.5 for x in the given inequality.

x^{2} + 5x + 6 = (-2.5)^{2} + 5(-2.5) + 6

x^{2} + 5x + 6 = 6.25 - 12.5 + 6

x^{2} + 5x + 6 = -0.25 <** **0

The interval [-3, -2] does not satisfy the given inequality.

**Testing the interval [-2, ****+∞] : **

Take a random value in the interval [-2, +∞), say

x = 0

Substitute 0 for x in the given inequality.

x^{2} + 5x + 6 = (0)^{2} + 5(0) + 6

x^{2} + 5x + 6 = 0 - 0 + 6

x^{2} + 5x + 6 = 6 **≥**** **0

The interval [-2, +∞) satisfies the given inequality.

Hence, the solution is

(-∞, -3] U [-2, +∞)

(or)

x ≤ -3 or x ≥ -2

**Example 2 :**

Solve the quadratic inequality given.

-x² + 2x - 5 ≤ 0

**Solution : **

Both (-x^{2} +2x - 5) and x^{2} have same signs (negative). So, there is solution for the given inequality.

To know the type of solution, let us find the value of

b^{2} -4ac

b^{2} - 4ac = (2)^{2} - 4(-1)(-5)

b^{2} - 4ac = 4 - 20

b^{2} - 4ac = -16 < 0

Hence, the solution is

all real values

(or)

x ∈ R

**Example 3 : **

Solve the quadratic inequality given.

-x^{2} + 4 ≥ 0

**Solution : **

Here, (-x^{2} + 4) and x^{2} have different signs.

Let us find the (b^{2} -4ac).

b^{2} -4ac = (0)^{2} - 4(-1)(4)

b^{2} -4ac = 0 + 16

b^{2} -4ac = 16 ≥ 0

(-x^{2} + 4) and x^{2} have different signs and b^{2} -4ac ≥ 0.

According to the shortcuts explained above, we have interval solution for the given quadratic inequality.

Let us find the interval solution.

Let -x^{2} + 4 = 0 and solve for x.

- x^{2} + 4 = 0

Multiply each side by -1.

x^{2} - 4 = 0

x^{2} - 2^{2} = 0

(x + 2)(x - 2) = 0

x = -2 or x = 2

Let us mark x = -2 and x = 2 on the real number line.

**Testing the interval (-∞, -2] : **

Take a random value in the interval (-∞, -2], say

x = -3

Substitute -3 for x in the given inequality.

-x^{2} + 4 = -(-3)^{2} + 4

-x^{2} + 4 = -9 + 4

-x^{2} + 4 = -5 < 0

The interval (-∞, -2] does not satisfy the given inequality.

**Testing the interval [-2, 2] : **

Take a random value in the interval [-2, 2], say

x = 0

Substitute 0 for x in the given inequality.

-x^{2} + 4 = -(0)^{2} + 4

-x^{2} + 4 = 4 ≥ 0

The interval [-2, 2] satisfies the given inequality.

**Testing the interval [2, ****+∞] : **

Take a random value in the interval [2, +∞), say

x = 3

Substitute 3 for x in the given inequality.

-x^{2} + 4 = -(3)^{2} + 4

-x^{2} + 4 = -9 + 4

-x^{2} + 4 = -5 < 0

The interval [2, +∞) does not satisfy the given inequality.

Hence the solution is

[-2, 2]

(or)

-2 ≤ x ≤ 2

**Example 4 : **

Solve the quadratic inequality given.

x^{2} + 4x +7 ≤ 0

**Solution : **

Here, (x^{2} + 4x + 7) and x^{2} have different signs.

Let us find the value of (b^{2} -4ac).

b^{2} -4ac = (4)^{2} - 4(1)(7)

b^{2} -4ac = 16 - 28

b^{2} -4ac = - 12 < 0

(-x^{2} + 4) and x^{2} have different signs and b^{2} -4ac < 0.

As per the shortcuts explained above, there is no solution for the given inequality.

After having gone through the stuff above, we hope that the students would have understood, how to solve quadratic inequalities algebraically.

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