SOLVING QUADRATIC EQUATIONS BY FACTORIZATION

Quadratic equations have the general form ax² + bx + c = 0, provided that a ≠ 0. Here are some examples of quadratic equations:

x² – 4x – 5 = 0          [a = 1, b = -4, c = -5]
-2x² – 3x + 9 = 0      [a = -2, b = -3, c = 9]
4x² + 7x = 0              [a = 4, b = 7, c = 0]
9x² – 16 = 0              [a = 9, b = 0, c = -16]

 

The number of roots of a quadratic equation can be determined by calculating the discriminant D = b² – 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)

x² – 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:

5² – 4⋅5 – 5 = 25 – 20 – 5 = 0

(Verify that D > 0.)

 

Example 2 (exactly one root)

x² – 6x + 9 = 0 has exactly one root, that is x = 3. Note that 3² – 6⋅3 + 9 = 9 – 18 + 9 = 0. No other value of x satisfies this equation. (Verify that D = 0.)

 

Example 3 (no real roots)

x² + 100 = 0 has no real roots. It is equivalent to x² = -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 6x² – 5x – 4 = 0.

Answer

Since 6x² – 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, 6x² – 5x – 4 is expressed as (3x – 4)(2x + 1). The process of “converting” 6x² – 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 ax² + 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 ax² + bx + c = 0, express bx as the sum of ux and vx.

In Example 4, it is performed as follows.

6x² – 5x – 4 = 0

⇔ 6x² + 3x + (-8x) – 4 = 0.

Step 4

Factorize (ax² + ux) and (vx + c).

In Example 4, the factorizations are performed as follows: 6x² + 3x = 3x(2x + 1) and (-8)x – 4 = -4(2x + 1)

Replace 6x² + 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 -2x² – 3x + 9 = 0.

Answer

To find the solution to the equation, we factorize -2x² – 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: -2x² – 6x + 3x + 9 = 0

Step 4

Factorize -2x² – 6x and 3x + 9 as follows.

-2x² – 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 ax² + c = 0 with ac ≤ 0 and a ≠ 0

Examples of quadratic equations of this form are:

x² – 16 = 0   [a = 1, b = 0, c = -16]
3x² – 25 = 0 [a = 3, b = 0, c = -25]
-2x² + 7 = 0 [a = -2, b = 0, c = 7]
-7x² = 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 -7x² = 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 ax² + c = 0 can be solved by factorization using the formula a² – b² = (a+b)(a-b). Look at some examples below.

 

Example 6

Find the roots of x² – 16 = 0

Answer

First note that x² – 16 = 0 ⇔ x² – 4² = 0.

By the formula a² – b² = (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. x₁ = -4 and x₂ = 4.

 

Example 7

Find the roots of 3x² – 25 = 0.

Answer

This quadratic equation can be rewritten as .

Using the formula a² – b² = (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 -2x² + 7 = 0

Answer

This quadratic equation can be rewritten as 7 – 2x² = 0, which is equivalent to (√7)² – (x√2)² = 0.

Using the formula a² – b² = (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 ax² + bx = 0 with a ≠ 0

Examples of quadratic equations of this form are:

5x² – 3x = 0      [a = 5, b = -3]
-x² + 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 5x² – 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 5x² – 3x = 0 are x = 0 or .

 

Example 10

Find the roots of -x² + 8x = 0

Answer

By factorization, -x² + 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 x₁ = 0 and x₂ = 8.

 

Example 11

Find the roots of 2x² = -3x.

Answer

The equation can be rewritten as 2x² + 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 x₁ = 0 and x₂ = -1½.