# CLASS-10QUADRATIC EQUATIONS SOLVING BY FACTORIZATION

Method of Solving a Quadratic Equation by Factorization

Step.1 Make the given equation free from factorization and radicals and put it into the standard form ax²+ bx + c = 0

Step.2)  Factorize (ax²+ bx + c) into two linear factors.

Step.3)  Put each linear factor equal to 0 (zero product rule).

Step.4) Solve these linear equations and get two roots of the given quadratic equations.

Example.1)  4x²+ 14x + 10 = 0

In above given equation is in form of ax² + bx + c = 0, here a = 4, b = 14, and c = 10. Now we have to factorize c = 14 in two part. So, the process is to multiply a X c = 4 X 10 = 40, then we have to do simple factorization of 40, and we find 40 = 2 X 2 X 2 X 5. So, 40 is the product of common factors 2, 2, 2, 5. and hence we have to find out addend of 14After bx there is positive (+) or plus (+) sign, so two obtained factor must be in addition form. Addend of 14, one must be 2 X 2 = 4, and another must be 2 X 5 = 10. So, two addend of 14 must be 4 & 10.

4 x² + 14x + 10 = 0

=>     4x² + (4 + 10) x + 10 = 0

=>     4x² + 4x + 10x + 10 = 0

=>     4x(x + 1) + 10(x + 1) = 0

=>     (x + 1) (4x + 10) = 0

=>     (x + 1) = 0  or  (4x + 10) = 0         [Zero Product Rule]

10            5

=>   x = -1, or  x = - ------ = - ----------

4             2

5

So, solution set = { -1, - ------ }         (Ans.)

2

Example.2)   5x² = 15x

In above given equation is in form of ax²+ bx + c = 0, here a = 5, b = -15, and c = 0. We not need to factorize c , because c = 14.

5x² = 15x

=>     5x² - 15x = 0

=>     5x (x – 3) = 0

=>     5x = 0  or   (x – 3) = 0           [Zero product rule]

=>     x = 0  or  x = 3

So,  Solution set = {0, 3}       (Ans.)

Example.3) Solve : 9x² – 22x + 8 = 0,  when (i)  x ϵ N, (ii)  x ϵ Q

Ans.)   9x²– 22x + 8 = 0

In above given equation is in form of ax²+ bx + c = 0, here a = 9, b = (-) 22, and c = 8. Now we have to factorize c = 22 in two part. So, the process is to multiply a X c = 9 X 8 = 72, then we have to do simple factorization of 40, and we find 72 = 3 X 3 X 2 X 2 X 2. So, 72 is the product of common factors 3, 3, 2, 2, 2, and hence we have to find out addend of 22. After bx there is positive (+) or plus (+) sign, so two obtained factor must be in addition form. Addend of 22, one must be 3 X 3 X 2 = 18, and another must be 2 X 2 = 4. So, two addend of 22 must be 18 & 4.

9x²+ (–) 22x + 8 = 0

=>    9x²+ (-) (18 + 4)x + 8 = 0

=>    9x²- 18x – 4x + 8 = 0

=>    9x (x – 2) – 4 (x – 2) = 0

=>    (x – 2) (9x – 4) = 0

=>    (x – 2) = 0  or  (9x – 4) = 0           [Zero product rule]

=>     x = 2,   or   x = 4/9

When, x ϵ N,

4

As,  2 ϵ N  and ------ ϵ  N, so the solution set = {2} ………………..(i)

9

When, x ϵ Q

4

As,  2 ϵ Q, and ------ ϵ Q

9

4

So, the solution set is = {2, -----}   …………………………(ii)

9

Example.4)  Solve 3a²x² + 11abx + 6b² = 0

Ans.)     3a²x²+ 11abx + 6b²= 0

In above given equation is in form of ax²+ bx + c = 0, here a = 3a², b = 11ab, and c = 6b². Now we have to factorize c = 11ab in two part. So, the process is to multiply a X c = 3a² X 6b² = 18a²b², then we have to do simple factorization of 18a²b², and we find 18a²b² = 3 X 3 X 2 X a X a X b X b. So, 18a²b² is the product of common factors 3, 3, 2, a, a, b, b,  and hence we have to find out addend of 18a²b². After bx there is positive (+) or plus (+) sign, so two obtained factor must be in addition form. Addend of 18a²b², one must be 3 X 3 X a X b = 9ab, and another must be 2 X a X b = 2ab. So, two addend of 11ab must be 9ab & 2ab.

3a²x² + 11abx + 6b² = 0

=>     3a²x² + (9ab + 2ab)x + 6b² = 0

=>     3a²x² + 9abx + 2abx + 6b² = 0

=>     3ax (ax + 3b) + 2b (ax + 3b) = 0

=>     (ax + 3b) (3ax + 2b) = 0                  [By Factorization]

=>     (ax + 3b) = 0  or  (3ax + 2b) = 0         [Zero Product Rule]

=>     x = - 3b/a   or   x = - 2b/3a

So, the solution set = {- 3b/a, - 2b/3a}        (Ans.)

Example.5)  Solve 12x²+ 5x  = 3

12x² + 5x = 3

=>     12x² + 5x – 3 = 0

=>     12x² + 5x + (- 3) = 0

In above given equation is in form of ax²+ bx + c = 0, here a = 12, b = 5, and c = 3. Now we have to factorize c = 5 in two part. So, the process is to multiply a X c = 12 X 3 = 36, then we have to do simple factorization of 36, and we find 36 = 3 X 3 X 2 X 2. So, 36 is the product of common factors 3, 3, 2, 2, and hence we have to find out addend of 36. After bx there is negative (-) or minus (-) sign, so two obtained factor must be in subtracted form. Addend of 36, one must be 3 X 3 = 9, and another must be 2 X 2 = 4. So, two addend of 11ab must be 9 & 2.

=>     12x² + 5x – 3 = 0

=>     12x² + (9 – 4)x – 3 = 0

=>     12x² + 9x – 4x – 3 = 0

=>     3x (4x + 3) – (4x + 3) = 0

=>     (4x + 3) (3x – 1) = 0                   [By Factorization]

=>     (4x + 3) = 0  or  (3x – 1) = 0        [Zero Product Rule]

=>      x = - 3/4  or  x = 1/3

So, the solution set = {- 3/4, 1/3}           (Ans.)