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 14. After 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.)
Your second block of text...