"Solving quadratic inequalities algebraically" is sometimes a difficult task for some students who study high school math.

Here we give you two shortcuts to easily solve any kind of quadratic inequalities algebraically.

In solving quadratic inequalities algebraically, shortcut 1 tells us whether there is solution for the given quadratic inequality.

**Other Cases :**

If both "ax² + bx + c " and "x²" have same signs, then there is solution for the given inequality.

If "ax² + bx + c " and "x²" have different signs,we have to find the value of "b² - 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.

To understand the above shortcuts clearly, let us look at some example problems on "Solving quadratic inequalities algebraically".

**Example 1 : **

Solve the quadratic inequality given.

**x² + 5x + 6 ≥ 0**

**Solution : **

Both "x² + 5x +6 " and "x²" have same signs (positive). So, there is solution for the given inequality.

To know the type of solution, let us find the value of "b² -4ac"

b² -4ac = 5² - 4(1)(6) = 25 - 24 = 1

We get,

b² -4ac = 1 --------->b² -4ac ≥ 0

Therefore, there is interval solution.

Let us find the interval solution.

Let x² + 5x + 6 = 0 and solve for "x".

x² + 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] : **

Let us take a value randomly from the interval (-∞, -3], say x = -4.

Plug x = -4, in the given inequality.

x² + 5x + 6 -----> (x+2)(x+3) -------> (-4+2)(-4+3) = (-2)(-1) = 2 ≥ 0.

For "x = -4", we get " **x² + 5x + 6 ****≥ 0**".

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

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

Let us take a value randomly from the interval [-3, -2], say x = -2.5

Plug x = -2.5, in the given inequality.

x²+5x+6 ----> (x+2)(x+3) -----> (-2.5+2)(-2.5+3) = (-0.5)(0.5) = -0.25 ≤ 0

For "x = -4", we get " **x² + 5x + 6 ****≤ ****0** ".

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

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

Let us take a value randomly from the interval [-2, +∞), say x = 0.

Plug x = 0, in the given inequality.

x² + 5x + 6 -----> (x+2)(x+3) ------> (0+2)(0+3) = (2)(3) = 6 ≥ 0.

For "x = 0", we get " **x² + 5x + 6 ****≥ 0**".

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

**Hence the solution is **

**(-∞, -3] U [-2, +∞)**

(or)

**x ≤ -3 or x ≥ -2**

Let us look at the next problem on "Solving quadratic inequalities algebraically"

**Example 2 : **

Solve the quadratic inequality given.

**-x² + 2x - 5 ****≤ 0 **

**Solution : **

Both "-x² +2x - 5 " and "x²" have same signs (negative). So, there is solution for the given inequality.

To know the type of solution, let us find the value of "b² -4ac"

b² -4ac = (2)² - 4(-1)(5) = 4 - 20 = -16

We get,

b² -4ac = -16 --------->b² -4ac < 0

**Hence, the solution is **

**all real values **

(or)

** x ∈ R **

Let us look at the next problem on "Solving quadratic inequalities algebraically"

**Example 3 : **

Solve the quadratic inequality given.

**-x² + 4 ****≥ 0**

**Solution : **

"-x² + 4 " and "x²" have different signs.

b² -4ac = 0 - 4(-1)(4) = 16 ≥ 0

"-x² + 4 " and "x²" have different signs and b² -4ac ≥ 0.

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

Let us find the interval solution.

Let -x² + 4 = 0 and solve for "x".

- x² + 4 = 0 -------> x² - 4 = 0 ------> (x+2(x-2) = 0 ------> x = -2, 2

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

**Testing the interval (-∞, -2] : **

Let us take a value randomly from the interval (-∞, -2], say x = -3.

Plug x = -3, in the given inequality.

-x² + 4 -----> -(x²-4) -------> -(9-4) = -5 < 0.

For "x = -4", we get " -**x² +4 <**** 0**".

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

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

Let us take a value randomly from the interval [-2, 2], say x = 0.

Plug x = 0, in the given inequality.

-x² + 4 -----> -(0-4) -------> -(-4) = 4 ≥ 0.

For "x = 0", we get " **-x² +4 ≥ 0**".

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

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

Let us take a value randomly from the interval [2, +∞), say x = 3.

Plug x = 3, in the given inequality.

-x² + 4 -----> -(9-4) -------> -5 < 0.

For "x = 3", we get " -**x² +4 <**** 0**".

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

**Hence the solution is**

**[-2, 2]**

(or)

**-2 ****≤ **x ≤ 2

Let us look at the next problem on "Solving quadratic inequalities algebraically"

**Example 4 : **

Solve the quadratic inequality given.

**x² + 4x +7 ****≤**** 0**

**Solution : **

"x² + 4x + 7 " and "x²" have different signs.

b² -4ac = 4² -4(1)(7) = 16 - 28 = 12 < 0

"-x² + 4 " and "x²" have different signs and b² -4ac < 0.

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

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Apart from the stuff and example problems explained above, if you want to know more about "solving quadratic inequalities algebraically", please click here.

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