Solving quadratic equations

Generally, to find the solution of a quadratic equation ax+ bx + c = 0 we use the quadratic formula, however there is another way to solve it: the factorization method.

Solving quadratic equations by the factorization method

There are two ways:

  1. We take the common factor of x and then look for the two values of it. This is used when c = 0.
  1. We convert the quadratic equation into a product of binomials. Then we find the value of x for each of them. To do this, we must find two numbers whose product is equal to "c" and whose sum is equal to b. This method is called simple factorization.

Examples

1. Solve the following quadratic equation x² + 3x = 0

Let's see that c = 0, therefore, we get a common factor of x:

x (x + 3) = 0

For this to be true, "x = 0" and "x + 3 = 0". Therefore, our first root of the equation is:

x1 = 0

Now,

x + 3 = 0

clearing x:

x = - 3

Thus:

x2 = – 3

Consequently, the roots of the equation are x1 = 0 and x2 = - 3; if we substitute one of them in the original equation we verify that it is a solution:

For x = 0:

x² + 3x = 0

(0) ² + 3 (0) = 0

0 = 0

For x = - 3:

(-3) ² + 3 (-3) = 0

9 - 9 = 0

0 = 0

2. Solve the following quadratic equation 6x² + 4x = 0

Let's see that c = 0, therefore, we get a common factor of x:

2x (3x + 2) = 0

For this to be true, "2x = 0" and "3x + 2 = 0". Therefore, our first root of the equation is:

2x1 = 0

x1 = 0 / 2

x1 = 0

Now,

3x + 2 = 0

clearing x:

3x = - 2

x = - 2/3

Thus:

x2 = – 2/3

Consequently, the roots of the equation are x1 = 0 and x2 = - 2/3; if we substitute one of them in the original equation we verify that it is a solution:

For x = 0:

6x² + 4x = 0

6 (0) ² + 4 (0) = 0

0 = 0

For x = - 2/3:

6 (- 2/3) ² + 4 (- 2/3) = 0

6 (4/9) - 8/3 = 0

3/8 - 8/3 = 0

0 = 0

3. Solve the following quadratic equation x² + x - 6 = 0

We convert the quadratic equation into a product of binomials:

(x + α) (x + β) = 0

We must look for two numbers α and β that when multiplied give the value of c and that when added together equal b. In this case, two numbers whose product is - 6 and whose sum is 1:

 (x + 3) (x - 2) = 0

Since 3 · (-2) = -6 and 3 + (-2) = 1

Now, for the above to be true, “(x + 3) = 0” and “(x - 2)” = 0. Therefore, our first root of the equation is:

x + 3 = 0

x = 0 - 3

x1 = – 3

Now,

x - 2 = 0

x = 0 + 2

x2 = 2

Consequently, the roots of the equation are x1 = - 3 and x2 = 2; If we substitute each one (separately) in the original equation, we will verify that they are a solution.

4. Solve the following quadratic equation x² + 6 = 7x

We order the equation:

x² - 7x + 6 = 0

We convert the quadratic equation into a product of binomials:

(x + α) (x + β) = 0

We must look for two numbers α and β whose product is 6 and their sum is - 7:

 (x - 6) (x - 7) = 0

Since (- 6) · (-1) = 6 and (- 6) + (- 1) = - 7

Now, for the above to be true, “(x - 6) = 0” and “(x - 7)” = 0. Therefore, our first root of the equation is:

x - 6 = 0

x = 0 + 6

x1 = 6

Now,

x - 7 = 0

x = 0 + 7

x2 = 7

Consequently, the roots of the equation are x1 = 6 and x2 = 7; If we substitute each one (separately) in the original equation, we will verify that they are a solution.