General form of a quadratic inequality, after moving all the expressions to one side of the inequality, is in one of the forms which are shown below.

$ax^2+bx+c > 0$   or   $ax^2+bx+c \geq 0$   or   $ax^2+bx+c < 0$   or   $ax^2+bx+c \leq 0$       (1)

When $a \neq 0$ and also $b, c \in \mathbb{R}$

The purpose of solving each of the inequality which is shown above is to find the set of all real numbers that we can substitute them instead of $x$ and the inequality is in correct form.

For example we claim that $x = 1$ is one of the answer of the $x^2 - \frac{1}{2} > 0$. Put 1 instead of all $x$ in the inequality, we would conclude that $1^2 - \frac{1}{2} > 0 \rightarrow \frac{1}{2} > 0$
which is always correct. So $x = 1$ is one of the answers of the inequality.

Now we will understand how to solve the (1) inequalities.

First, we need to look at the associated two-variable equation, $y = ax^2+bx+c$, and consider where $ax^2+bx+c$ is equal to zero:

$ax^2+bx+c = 0 \rightarrow a(x^2+\frac{b}{a}x+\frac{c}{a}) = 0 \rightarrow^{a \neq 0} x^2+\frac{b}{a}x+\frac{c}{a} = 0 \rightarrow$
$x^2+\frac{b}{a}x+\frac{c}{a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2} = 0 \rightarrow (x + \frac{b}{2a})^2 - \frac{b^2 - 4ac}{4a^2} = 0 \rightarrow$
$(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2} \rightarrow x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} \rightarrow x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \rightarrow$
$x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

This says that the quadratic crosses the x-axis at $x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ and at $x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

These zeroes divide the number line into three intervals:

$(-\infty, x_1)$   ,   $[x_1,x_2]$   ,   $(x_2,+\infty)$

With this assumption that $x_1 < x_2$.

Now let $\Delta = b^2 - 4ac$.

We can consider three different cases as follow:

1. $\Delta > 0$
2. $\Delta = 0$
3. $\Delta < 0$

Case 1:
If $\Delta > 0$

then $ax^2+bx+c$ has two different roots $(x_1 \neq x_2)$.
Now if $a>0$ then its graph is like "Figure a".
If $a<0$ then its like "Figure b". So if $a>0$ and also if we have $ax^2+bx+c \geq 0 (ax^2+bx+c > 0)$ then the set of the answer is:
$(-\infty, x_1] \cup [x_2, +\infty)$       $((-\infty, x_1) \cup (x_2, +\infty))$
And if we have $ax^2+bx+c \leq 0 (ax^2+bx+c < 0)$ then the set of solutions is:
$[x_1,x_2]$         $((x_1,x_2))$
On the other hand if $a < 0$ and also if we have $ax^2+bx+c \geq 0 (ax^2+bx+c > 0)$ then the set of the answer is:
$[x_1,x_2]$         $((x_1,x_2))$
And if we have $ax^2+bx+c \leq 0 (ax^2+bx+c < 0)$ then the set of the answer is:
$(-\infty, x_1] \cup [x_2, +\infty)$       $((-\infty, x_1) \cup (x_2, +\infty))$
Case 2:
If $\Delta = 0$

then $ax^2+bx+c = 0$ has only one root $(x_1 = x_2 = \frac{-b}{2a})$. Now if $a>0$ then its graph is like "Figure c" and if $a<0$ then its like "Figure d".

If $ax^2+bx+c \geq 0$ and a > 0 then solutions are all real numbers.
If a > 0 and $ax^2 + bx + c > 0$ (or a < 0 and $ax^2 + bx + c < 0$) then the set of solutions is $R - \{\frac{-b}{2a}\}$(All numbers except $\{\frac{-b}{2a}\}$)

It’s clear that if a < 0 and $ax^2 + bx + c \geq 0$(or a > 0 and $ax^2 + bx + c \le 0$) then the inequality has one solution only and it is $\{\frac{-b}{2a}\}$ and also
if a < 0 and $ax^2 + bx + c > 0$(or a > 0 and $ax^2 + bx + c < 0$) then the inequality has no real solutions.
Case 3:
If $\Delta < 0$

then $ax^2+bx+c=0$ has no root. Now if $a>0$ then its graph is like "Figure e" and if $a<0$ then its like "Figure f".

If $ax^2 + bx + c \ge 0$ or $ax^2 + bx + c > 0$ and $a > 0$(or $ax^2 + bx + c < 0$ and $a < 0$) then solutions are all real numbers.
It’s clear that if $ax^2 + bx + c > 0$ and $a < 0$(or $ax^2 + bx + c < 0$ and $a > 0$) then the quadratic inequality has no real solutions.

Example 1: Find the set of the answer of $x^2 + 3x - 10 > 0$.

Solution: According to what we said above

$\Delta = b^2 - 4ac \rightarrow \Delta = 3^2 - 4(1)(-10) = 9 + 40 = 49 \rightarrow \Delta > 0 \rightarrow x^2 + 3x - 10$ has two different roots which is:

$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} = \frac{-3 + \sqrt{49}}{2 \times 1} = \frac{-3 + 7}{2} = \frac{4}{2} = 2$

$x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-3 - \sqrt{49}}{2 \times 1} = \frac{-3 - 7}{2} = \frac{-10}{2} = -5$

On the other hand $a > 0$, so the set of the answer of the inequality according to case 1 and "Figure a" is:

$(-\infty, -5) \cup (2, +\infty)$

Example 2: Find the set of the answer of $x^2 + 5x - 6 \geq 0$.

Solution: According to what we said above

$\Delta = b^2 - 4ac \rightarrow \Delta = 5^2 - 4(1)(-6) = 25 + 24 = 49 \rightarrow \Delta > 0$
$x^2 + 5x - 6 = 0$ has two different roots which is:

$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} = \frac{-5 + \sqrt{49}}{2 \times 1} = \frac{-5 + 7}{2} = \frac{2}{2} = 1$

$x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-5 - \sqrt{49}}{2 \times 1} = \frac{-5 - 7}{2} = \frac{-12}{2} = -6$

On the other hand $a > 0$, so the set of the answer of the inequality according to case 1 and "Figure a" is:

$(-\infty, -6] \cup [1, +\infty)$

Example 3: Find the set of the answer of $x^2 - 2x + 1 \geq 0$.

Solution: According to what we said above

$\Delta = b^2 - 4ac \rightarrow \Delta = (-2)^2 - 4(1)(1) = 4 - 4 = 0 \rightarrow \Delta = 0$
$x^2 - 2x + 1 = 0$ has has only one root which is:

$x_1 = x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-2) \pm \sqrt{0}}{2 \times 1} = \frac{2 \pm 0}{2} = \frac{2}{2} = 1$

On the other hand $a > 0$, so the set of the answer of the inequality according to case 1 and "Figure c" is: $\mathbb{R}$

Example 4: Find the set of the answer of $x^2 - 2x + 1 < 0$.

Solution: According to what we said above

$\Delta = b^2 - 4ac \rightarrow \Delta = (-2)^2 - 4(1)(1) = 4 - 4 = 0 \rightarrow \Delta = 0$
$x^2 - 2x + 1$ has has only one root which is:

$x_1 = x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-2) \pm \sqrt{0}}{2 \times 1} = \frac{2 \pm 0}{2} = \frac{2}{2} = 1$

On the other hand $a > 0$, so $x^2 - 2x + 1 \geq 0$ therefore according to all the things we said here and also to the graph of the previous example the $x^2 - 2x + 1 < 0$ has no answer in real numbers.

Example 5: Find the set of the answer of $-x^2 - 7x + 8 < 0$.

Solution: According to what we said above

$\Delta = b^2 - 4ac \rightarrow \Delta = (-7)^2 - 4(-1)(8) = 49 + 32 = 81 \rightarrow \Delta > 0$
$-x^2 - 7x + 8$ has two different roots which is:

$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} = \frac{-(-7) + \sqrt{81}}{2 \times (-1)} = \frac{7 + 9}{-2} = \frac{16}{-2} = -8$

$x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-(-7) - \sqrt{81}}{2 \times (-1)} = \frac{7 - 9}{-2} = \frac{-2}{-2} = 1$

On the other hand $a > 0$, so the set of the answer of the inequality according to case 1 and "Figure a" is:

$(-\infty, -8) \cup (1, +\infty)$

Example 6: Find the set of the answer of $x^2 + 1 > 0$.

Solution: According to what we said above

$\Delta = b^2 - 4ac \rightarrow \Delta = 0^2 - 4(1)(1) = 0 - 4 = -4 \rightarrow \Delta < 0$
$x^2 + 1 = 0$ has no root in real numbers. On the other hand $a > 0$, so the set of solutions of $x^2 + 1 > 0$ is $\mathbb{R}$

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