  How to solve quadratic inequalities: examples and their solutions.

## Example 1: Solve x2 - 3x - 10 ≤ 0 Factor x2 - 3x - 10.

Find a pair of numbers
whose product is the constant term [-10]
and whose sum is the middle term's coefficient [-3].

The constant term is (-).
So the signs of the numbers are different:
one is (+), and the other is (-).

(-1, 10) and (-2, 5)
are not the right numbers.

[-10] = -5⋅2
-5 + 2 = [-3]
So -5 and 2 are the right numbers.

Use -5 and +2
to write a factored form:
(x - 5)(x + 2) ≤ 0.

Find the zeros of (x - 5)(x + 2) ≤ 0.

1) x - 5 = 0
So x = 5.

2) x + 2 = 0
So x = -2.

So the zeros are x = 5, -2.

Solving a quadratic equation by factoring

Roughly draw y = (x - 5)(x + 2)
on the x-axis.

Use the zeros 5 and -2.

The inequality sign shows that
y = (x - 5)(x + 2) ≤ 0.

So color the function below the x-axis,
including the zeros: -2, 5.

Then the range of x is
-2 ≤ x ≤ 5.

## Example 2: Solve x2 - 42 > 0 16 = 42

x2 - 42 = (x + 4)(x - 4)

Factor the difference of two squares (a2 - b2)

Find the zeros of (x + 4)(x - 4) > 0.

1) x + 4 = 0
So x = -4.

2) x - 4 = 0
So x = 4.

So the zeros are x = ±4.

Solving a quadratic equation by factoring

Roughly draw y = (x + 4)(x - 4)
on the x-axis.

Use the zeros ±4.

The inequality sign shows that
y = (x + 4)(x - 4) > 0.

So color the function above the x-axis,
excluding the zeros: ±4.

Then x < -4 or x > 4.

## Example 3: Solve -x2 + 10x - 25 ≥ 0 To make the coefficient of x2 (+),
multiply both sides by -1.

Don't forget to change
the order of the inequality sign.

Linear inequality (One variable)

-10x = -2⋅5⋅x
+25 = +52

x2 - 2⋅5⋅x + 52 = (x - 5)2

Factor a perfect square trinomial (a2 ± 2ab + b2)

Find the zeros of (x - 5)2 ≤ 0.

x - 5 = 0
So x = 5.

This is the zero.

Roughly draw y = (x - 5)2
on the x-axis.

Use the zero 5.

The inequality sign shows that
y = (x - 5)2 ≤ 0.

So color the function below the x-axis,
including the zero: 5.

Only x = 5 is in the range.

So x = 5 is the answer.