Solution of Quadratic Equation by Factorization Method

Examples:

Solve the following quadratic equations by the factorization methods.

• m² = 11m + 42
• 2x² – x – 21 =0
• 2p² + 11p = 30.

Solutions.

(1 ) m² = 11m + 42

then m² – 11m – 42 =0

1^st step:  Find the product of the first and last  terms :

m² x – 4² =-42m²

2^nd step:find two terms such that their product is -42m2 and their sum is -11m ( the middle term).

Factors of -42m² sum of factor s

(a)  + 3m and – 14m                         – 11m

(b)  – 6m and +7m                             + 11m

( c) -6m and + 7m                             +m

(d) +6m and -7m                               -m

Out of these, only (a) gives the required result.

3^rd step:  Replace -11 in the given expression by +3m and -14m.  then factorise by grouping:

m² -11m -42 = 0.

M² + 3m  – 14m – 42 =0

M(m+ 3) (m – 14) = 0

M + 3 = 0 or m – 14 =0

i.e m = 0 -3, or m = 0 + 14.

M  = -3 or m = 14.

2. 2x² –x -21 =0

1^st Step:  Product  of first and last terms = 2x² x -21

= – 42x²

2^nd step:  Find two terms such that their product is 42x² and their sum is –x (the middle term).

Factors of-42x²                  sum of factors

(a) + 2x and -21x                               -19x

(b) -2x and 21x                   +19x

( c)-3x and + 14x                               + 11x

(d) + 3x and -14x                               -11x

( e) +6x and -7x                 -x

(f)   -6x and + 7x                +x

out of these, only (e)gives the required result.

3^rd step:  Replace –x in the given expression by +6x and -7x. then factorize by grouping

2x²  – x -21  =0

2x²  +6x -7x – 21 = 0

x ( x + 3)  -7 ( x + 3 ) = 0

( x + 3) ( x-7)  =0

either x + 3  = 0 or 2x -7 =0

i.e x = 0-3  or 2x = 7

x = -3  or x =⁷/2

x = -3  or x = 3 ½ .

3. 2p² + 11p  = 30.

2p² + 11p – 30 = 0

1^st step: Product of the first and last terms

2p² x -30  =- 60p²

2^nd step: find two terms such that their products is -60p² and their sum is +11p (the middle term )

Factors of +11                    sum of factors

(a) + 6p and – 10p            -4p

(b) -15p and +4p               +4p

( c) -15p and +4p              -11p

(d) +15p and -4p               +11p.

Out of these, only (d) gives the more required solution. Replace + 11p by +15p and -4p in the  given expression .

2p² + 15p -4p -30 = 0

p ( 2p + 15) – 2 (2p + 15) = 0

(2p + 15) (p – 2) = 0

either 2p+ 15  = 0 or p -2 = 0

2p  = 0 – 15 or p = 0 + 2

2p = -15 or p = 2

p = -15 or p = 2

p = –¹⁵/2 or p = 2

p = -7 ½  or p = 2.

EVALUATION (USE THE BOX AT THE BOTTOM TO POST YOUR ANSWER FOR DISCUSSION AND APPRAISAL).

Solve the following quadratic equations by factorization method.

• 4e² – 20 e + 25 = 0
• 4a² – 11a = 3

ASSIGNMENT  (USE THE BOX AT THE BOTTOM TO POST YOUR ANSWER FOR DISCUSSION AND APPRAISAL).

1. If x² – 10 x – 24 = 0, then x = 12 or ……………

(a) -3                      (b) -2                     (c)  -1    (d) 1     (e) 2

2. if x² + Kx + 16/9 is a  perfect square, then K = ………………..

(a) ¹/3     (b)  ²/3       (c)  ⁴/2            (d) ⁸/3            (e)  ¹⁶/3

3. What is the sum of the roots of the equation

x² -3x +2 =0?

(a) – 3                    (b) -1   ( c) 2    (d )3    (e)   5

4. Find the roots of the equation

x² + 12x – 28 = 0

The greater of the two roots is ………….

(a) -14         (b) -2        (c ) 2(d)7   (e ) 14.

5. What are the factors of 6×2 + x – 12?

A.(2x+ 3)( x + 6)              B. (2x +3) ( 2x – 3)        C. (2x +3)(3x + 4)   D. (2x + 3) (3x -4)

1. (2x + 3 ) ( 4x -9).