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Math Lesson 9.5.1 - Solving Quadratic Equations by Factorizing

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Welcome to our Math lesson on Solving Quadratic Equations by Factorizing, this is the first lesson of our suite of math lessons covering the topic of Quadratic Equations, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Recalling Quadratic Equations. Solving Quadratic Equations by Factorizing

From tutorials 6.4, it is known that a quadratic equation is a second - order equation with one variable that has a general form

ax2 + bx + c = 0

It is called a 'second - order equation' because its variable x is in the second power. On the other hand 'with one variable' means that only the variable x is unknown; the other letters represent known numbers, where a and b are called coefficients while c is a constant.

For example,

5x2 - 3x - 2 = 0

is a quadratic equation, where a = 5, b = -3 and c = -2.

Solving a quadratic equation means finding the value(s) of the variable x (we call them 'roots') for which the equation becomes true. For example, x = 1 is a root (solution) of the above equation as

5 ∙ 12 - 3 ∙ 1 - 2
= 5 - 3 - 2
= 0

gives a true result, while x = 3 is not a root of the same equation, as

5 ∙ 32 - 3 ∙ 3 - 2
= 5 ∙ 9 - 3 ∙ 3 - 2
= 45 - 9 - 2
= 34 ≠ 0

In tutorials 6.4, we explained just one method for solving quadratic equations, which consists of factorizing the left side of the original equation in the form

(px - m)(qx - n) = 0

where

p ∙ q = a
m ∙ n = c

and

- np - mq = b

are relations involving new coefficients m, n, p and q to be found mostly by guessing. For this reason, it is not very appropriate trying to solve quadratic equations by this kind of factorisation, although here we are going to recall this method again, in order to make it join the rest of the methods explained later on in this tutorials.

Example 1

Find by factorisation the roots of the quadratic equation

x2 - 4x + 3 = 0

Solution 1

We have a = 2, b = -4 and c = 3.

Since we have to express the equation in the form

(px - m)(qx - n) = 0

where

p ∙ q = a
m ∙ n = c

and

- np - mq = b

we have

p ∙ q = 1
m ∙ n = 3

and

- np - mq = - 4

We, therefore, have the following combinations:

p = 1;
q = 1;
m = - 3, - 1, 1, or 3;
n = - 1, - 3, 3 or 1

Considering all possible combinations for all four unknowns in the three given equations, we obtain p = 1, q = 1, m = 3 and n = 1.

Therefore, our original equation is factorised as

(x - 3)(x - 1) = 0

Its roots must be such that to make the brackets zero. Thus, the first root is

x1 - 1 = 0
x1 = 1

and the second root is

x2 - 3 = 0
x2 = 3

Proof: Expanding the last equation yields

(x - 3)(x - 1) = 0
x ∙ x + x ∙ ( - 1) - 3 ∙ x - 3 ∙ ( - 1) = 0
x2 - x - 3x + 3 = 0
x2 - 4x + 3 = 0

which is the same as the original equation.

As you see, this method is too long and energy - consuming, as it needs to consider many possible combinations until you get the correct solution. Perhaps you have to restart from the beginning when you realize that a certain combination does not result as correct during the way of solution.

Another shortcoming of the factorization method consists in the fact that not all coefficients or constants may be integers. Or they may be integers but none of the combinations meets the description. So far we have considered only quadratic equations which have two distinct roots, but it may occur that the equation has one or no roots, as we are going to see later. Therefore, the factorization method is very limited and not suitable to use in most cases.

More Quadratic Equations Lessons and Learning Resources

Equations Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
9.5Quadratic Equations
Lesson IDMath Lesson TitleLessonVideo
Lesson
9.5.1Solving Quadratic Equations by Factorizing
9.5.2Special Cases of Quadratic Equations
9.5.3Solving Quadratic Equations by Factorization Part Two
9.5.4Solving a Quadratic Equation by Completing the Square

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