Quadratic equations have the general form ax^{2} + bx + c = 0, provided that a ≠ 0. Here are some examples of quadratic equations:

x^{2} – 4x – 5 = 0 [a = 1, b = -4, c = -5]

-2x^{2} – 3x + 9 = 0 [a = -2, b = -3, c = 9]

4x^{2} + 7x = 0 [a = 4, b = 7, c = 0]

9x^{2} – 16 = 0 [a = 9, b = 0, c = -16]

The number of roots of a quadratic equation can be determined by calculating the discriminant *D* = b^{2} – 4ac. If D > 0 then the quadratic equation has two distinct roots. If *D* = 0, it has exactly one root. If *D* < 0, it has no real roots. By real roots, we mean the roots which are real numbers.

**Example 1 (two distinct roots)**

x^{2} – 4x – 5 = 0 has two distinct solutions, namely x = -1 or x = 5. To prove it, substitute -1 for x in the equation. This results in:

(-1)^{2} -4(-1) – 5 = 1 + 4 – 5 = 0

Similarly, substituting x = 5 into the equation gives:

5^{2} – 4⋅5 – 5 = 25 – 20 – 5 = 0

(Verify that *D* > 0.)

**Example 2 (exactly one root)**

x^{2} – 6x + 9 = 0 has exactly one root, that is x = 3. Note that 3^{2} – 6⋅3 + 9 = 9 – 18 + 9 = 0. No other value of x satisfies this equation. (Verify that *D* = 0.)

**Example 3 (no real roots)**

x^{2} + 100 = 0 has no real roots. It is equivalent to x^{2} = -100. But this is impossible because the square of any real number cannot be negative.

**How to find these roots?**

There are several methods, but this post deals with the use of factorization. If a quadratic equation has two distinct roots, the solution to the equation can be determined by applying (+++) below.

pq = 0 **if and only if** p = 0 or q = 0 ………………………………………………………………………………………………………………………………… (+++)

In (+++), p and q are any real numbers. From (+++), it can be inferred that “if the product of two real numbers is zero, then at least one of the factor is zero.” It is impossible that p ≠ 0 and q ≠ 0 while pq = 0.

**Example 4**

Find the solution to the equation 6x^{2} – 5x – 4 = 0.

**Answer**

Since 6x^{2} – 5x – 4 = (3x – 4)(2x + 1), the quadratic equation can be rewritten as:

(3x – 4)(2x + 1) = 0 ……………………………………………………………………………………………………………………………………………………………………… (*)

The left-hand side of (*) is a product of two real numbers, i.e. (3x – 4) and (2x + 1). Since the product of these two is zero, according to (+++), 3x – 4 = 0 or 2x + 1 = 0. Note that:

3x – 4 = 0 ⇔

and

2x + 1 = 0 ⇔ .

In conclusion, the quadratic equation has two distinct roots, i.e. and .

When solving the quadratic equation in Example 4, 6x^{2} – 5x – 4 is expressed as (3x – 4)(2x + 1). The process of “converting” 6x^{2} – 5x – 4 to (3x – 4)(2x + 1) is called * factorization*. Below are the steps to factorize some given terms in an algebraic expression.

**To factorize ax ^{2} + bx + c = 0, perform the following steps**

Step 1

Multiply a by c. This yields ac.

In Example 4, a = 6 and c = -4. Thus, ac = 6(-4) = -24.

Step 2

Find two numbers u and v such that uv = ac and u + v = b.

In Example 4, b = -5.

Now let’s find two numbers u and v such that uv = -24 and u + v = -5.

It turns out that u = 3 and v = -8 meet the requirements.

Step 3

In the equation ax^{2} + bx + c = 0, express bx as the sum of ux and vx.

In Example 4, it is performed as follows.

6x^{2} – 5x – 4 = 0

⇔ 6x^{2} + 3x + (-8x) – 4 = 0.

Step 4

Factorize (ax^{2} + ux) and (vx + c).

In Example 4, the factorizations are performed as follows: 6x^{2} + 3x = 3x(2x + 1) and (-8)x – 4 = -4(2x + 1)

Replace 6x^{2} + 3x above by 3x(2x + 1) and (-8x) – 4 by -4(2x + 1). This gives:

3x(2x + 1) – 4(2x + 1) = 0

Step 5 (final step)

Factorize the result obtained in Step 4. Note that the terms have a common factor, i.e. (2x + 1).

This yields (2x + 1)(3x – 4) = 0.

**Example 5**

Find the solution to the equation -2x^{2} – 3x + 9 = 0.

**Answer**

To find the solution to the equation, we factorize -2x^{2} – 3x + 9 using the method as described above.

Step 1

Calculate the product ac.

In this example, a = -2 and c = 9. So, ac = -2⋅9 = -18.

Step 2

In this example, b = -3.

Find two numbers u and v such that uv = -18 and u + v = -3. We get u = -6 and v = 3.

Step 3

The quadratic equation can be expressed as: **-2x ^{2} – 6x + 3x**

**+ 9 = 0**

Step 4

Factorize -2x^{2} – 6x and 3x + 9 as follows.

-2x^{2} – 6x = -2x(x + 3)

3x + 9 = 3(x + 3)

Consequently, the quadratic equation can be rewritten as -2x(x+3) + 3(x+3) = 0.

Step 5

Factorizing the terms on the left-hand side of the equation just-obtained, we get:

(x+3)(-2x+3) = 0 ………………………………………………………………………………………………………………………………………………………………………. (**)

Now, we seek the roots of the equation.

Consider (**). According to (+++), the possibilities are (x+3) = 0 or (-2x+3) = 0. As x + 3 = 0 ⇔ x = -3 and -2x + 3 = 0 ⇔ , we can conclude that the roots of the equation are and .

**Special case: quadratic equation of the form ax ^{2} + c = 0 with ac ≤ 0 and a ≠ 0**

Examples of quadratic equations of this form are:

x^{2} – 16 = 0 [a = 1, b = 0, c = -16]

3x^{2} – 25 = 0 [a = 3, b = 0, c = -25]

-2x^{2} + 7 = 0 [a = -2, b = 0, c = 7]

-7x^{2} = 0 [a = -7, b = 0, c = 0]

In the four examples above, b = 0 and ac ≤ 0. The quadratic equations of this type always have two distinct roots. Furthermore, if c = 0 then the only solution to the equation is x = 0. For example, the equation -7x^{2} = 0 above has exactly one root, that is x = 0. Now we’ll look more closely at the case in which ac < 0.

If ac < 0, the equation ax^{2} + c = 0 can be solved by factorization using the formula a^{2} – b^{2} = (a+b)(a-b). Look at some examples below.

**Example 6**

Find the roots of x^{2} – 16 = 0

**Answer**

First note that x^{2} – 16 = 0 ⇔ x^{2} – 4^{2} = 0.

By the formula a^{2} – b^{2} = (a+b)(a-b), the equation can be expressed as (x + 4)(x – 4) = 0

This gives two possibilities, namely x + 4 = 0 or x – 4 = 0. [See (+++).]

Note that x + 4 = 0 ⇔ x = – 4 and x – 4 = 0 ⇔ x = 4

Thus, the quadratic equation has two distinct roots, i.e. x_{1} = -4 and x_{2} = 4.

**Example 7**

Find the roots of 3x^{2} – 25 = 0.

**Answer**

This quadratic equation can be rewritten as .

Using the formula a^{2} – b^{2} = (a+b)(a-b), it follows that .

This gives two possibilities, namely x√3 + 5 = 0 or x√3 – 5 = 0. [See (+++).]

Note that x√3 + 5 = 0 ⇔ and x√3 – 5 = 0 ⇔ .

So, this quadratic equation has two distinct roots, namely and .

**Example 8**

Find the solutions to -2x^{2} + 7 = 0

**Answer**

This quadratic equation can be rewritten as 7 – 2x^{2} = 0, which is equivalent to (√7)^{2} – (x√2)^{2} = 0.

Using the formula a^{2} – b^{2} = (a+b)(a-b), we obtain the equation below.

(√7 + x√2)(√7 – x√2) = 0

By (+++), there are two possibilities, namely 7 + x√2 = 0 or 7 – x√2 = 0.

Since 7 + x√2 = 0 ⇔ and 7 – x√2 = 0 ⇔ , we can conclude that the desired solutions are or .

**Special case: quadratic equations of the form ax ^{2} + bx = 0 with a ≠ 0**

Examples of quadratic equations of this form are:

5x^{2} – 3x = 0 [a = 5, b = -3]

-x^{2} + 8x = 0 [a = -1, b = 8]

The left-hand side of such quadratic equation is easily factorized into x(ax + b) so that one of its roots must be zero. The other root is .

**Example 9**

Find the solutions to 5x^{2} – 3x = 0.

**Answer**

Factorize the left-hand side of the equation. Then, we can rewrite the equation as x(5x – 3) = 0.

This results in two possibilities, namely x = 0 or 5x – 3 = 0. The later equation is equivalent to . Consequently, the solutions to 5x^{2} – 3x = 0 are x = 0 or .

**Example 10**

Find the roots of -x^{2} + 8x = 0

**Answer**

By factorization, -x^{2} + 8x = = x(-x + 8). Thus, the equation is equivalent to x(-x + 8) = 0.

By (+++), there are two possibilities, that is x = 0 or -x + 8 = 0. As -x + 8 = 0 ⇔ x = 8, we can conclude that the roots of the equation are x_{1} = 0 and x_{2} = 8.

**Example 11**

Find the roots of 2x^{2} = -3x.

**Answer**

The equation can be rewritten as 2x^{2} + 3x = 0. By factorizing the left-hand side of the equation, we then have x(2x + 3) = 0. By (+++), there are two possibilities, namely x = 0 or 2x + 3 = 0. Note that 2x + 3 = 0 ⇔ x = -1½. Thus, the roots of the equation are x_{1} = 0 and x_{2} = -1½.