Categories
Mathematics

Matrices and Determinants

A quadratic equation is an expression of the form ax2 + bx + c = 0 in which a, b & c are numerals; and also the highest power of x is 2 & that the power of x will neither be fractions nor negatives. Quadratic equations can be solved using the method of factorization, completing the square, quadratic formula& graphical method

Steps in solving quadratic equation: (1)examine the middle term whose power of x is 1. (2) Find the product of the first & last term. (3) Find two terms whose sum is equal to the middle term & product is equal to the value of the product of the first & last term (4) Replace the middle term by two the two terms in step 3. (5) Factorize the first two & last two terms (6) equate the linear factors to zero to find the value of x.

Example – Solve by factorization: X2 + 7X + 10 = 0

Solution

X2 + 7X + 10 = 0

X2 +2X + 5X + 10 = 0

X(X  + 2) + 5(X + 2) = 0

(X+2) (X + 5) = 0

X + 2 = 0

X = -2        OR

X + 5 = 0

X = -5

Hence X = -2 or -5

Factorization of Quadratic Expressions.

Factorize the following:

  1. 6a2 + 15a + 9
  2. 6a2 – 19ax  – 36x2
  3. (5x – 1) ( x – 3)  – ( x – 5)  ( x – 3)
  4. 35 – 2b  – b2
  5. x2 – y2 + ( x + y ) 2
  6. 25a2  – 4  ( a – 2b ) 2.

Solutions.

6a2 + 15x  + 9.

Since 3 is a common factor to all the terms, first take out 3 as the common factor:

6a2 + 15a  + 9  = 3 (2a2+ 5a + 3)

= 3 (2a2 + 2a + 3a + 3)

= 3 (2a (a+1)  + 3 (a+1)

= 3 (  (a+1)   (2a+ 3)).

Hence.

6a2 + 15a + 9 = 3   (a+ 1)  (2a + 3)

= 3 (a+1)  ( 2a + 3)

  1. 6a2 – 19ax – 36x2

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

6a2 x -36x2 = -216a2x2

2nd step:  Find two terms such that their products is – 216a2x and their sum is -19ax ( the middle term).

Factors of -216a2x2 sum of factors .

  1. +27ax and -8ax + 19ax
  2. -27ax and +8ax – 19ax
  3. +9ac and -24ax    – 15ax
  4. -9x and 24ax  + 15ax.

Of these only b gives the required result.

3rd step:  Replace  -19ax in the given expression by -27ax and 8ax then factorize by grouping:

6a2 – 19ax – 36x2

= 6a2 – 27ax  + 8ax – 36x2

= 3a (2a – 9x)  + 4x (2a-9x)

=(2a – 9x ) ( 3a + 4x)

hence,

6a2 – 19ax – 36x2 = (2a -9x) (3a + 4x)

  1. (5x – 1 (x -3) – (x -5)(x – 3)

= (x – 3)   ( 5x -1)  – (x – 5)

= ( x -3)    ( 5x – 1  – x + 5 )

=  ( x – 3)  ( 5x  – x + 5 – 1)

=  ( x – 3)   ( 4x + 4)

=  ( x – 3)  + ( x + 1)

= ( x -3)  4(x+1)

=  4(x-3) (x+1)

  1. 35 – 2b –b2

or  – b2 – 2b + 35

= -b2 – 7b  + 5b  + 35

= -b (b+7)  + 5 (b + 7)

=  (b+7)  ( -b + 5)

=  (b+7)  ( 5-b)

or  (7+b)   ( 5 – b)

  1. x2 – y2 + (x + y)2

Since

X2 – y2  = (x)2 – (y)2

= ( x + y)  ( x –y)

then x2 – y2 + (x + y)2  = (x + y)(x-y) + (x+ y)2

= (x +y)( x-y + (x + y)

= (x + y  ( x –y  + x + y)

= ( x + y ) ( x + x

= (x + y)  (2x)

(x +y ) (2x)  = 2x ( x +y)

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

Factorize the following

  1. m2 – 15mm – 54n2
  2. 8a2 – 18b2
  3. If 17x = 37 52  – 3562

Find the value of x

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