TUTORS SOLUTION ON QUADRATIC EQUATION USING FACTORIZATION METHOD.

Solution

1. $a^{2}-3a+2=0$

This means that the sum root is -3 and the product root is $2a^{2}$

The numbers such that their product gives $2a^{2}$ and their sum gives -3 is –a and 2a. Replace –3a the middle term of the equation .

$a^{2}-3a+2=0$ which is -3a, by –a and 2a as follows.

$a^{2}-3a+2=0$

$a^{2}-a-2a+2=0$

Group the above and factorize each brackets as follows.

$\left ( a^{2}-a \right )-\left (2a+2 \right )$

factorizing,we have;

a(a – 1) - 2(a - 1) = 0

In factorization of quadratic equation, once the terms of the two brackets are the same, then, your solving is correct otherwise, its wrong. Looking at too brackets, it contains like terms which signifies that we are correct.

Now, we factorize the LHS (Left Hand Side) of the equation;

a( a – 1 ) - 2( a - 1) = 0

The whole trick to this is: The terms put side the brackets while form a term in bracket while and one of any of the above brackets the contains like terms must be taken as follows.

( a – 2 )( a - 1)=0

Therefore , a - 2= 0 or a - 1 =0

This implies that a=2 or a=1

2. $x^{2}-x+2=0$

This means that the sum root is -x and the product root is 2x^2

The numbers such that that their product is 2x^2 and their sum gives -x is x and -2x. Replace –x, the middle term of the given equation $x^{2}-x+2=0$ by x and 2x as follows.

. $x^{2}-x+2=0$

$x^{2}+x-2x+2=0$

Group the above and factorize each brackets as follows.

$\left (x^{2}+x \right )-\left (2x+2 \right ) =0$

x(x +1) - 2(x +1) = 0

In factorization of quadratic equation, once the terms of the two brackets are the same, then, your solution is correct otherwise, it’s wrong. Looking at the two brackets, it contains like terms which signifies that we are correct.

Now, we factorize the LHS (Left Hand Side) of the equation;

x( x + 1 ) - 2( x + 1) = 0

The whole trick to this is: The terms outside side the brackets whill form a term in a bracket while one of the above brackets that contain like terms must be taken out as another factor as follows.

( x – 2 )( x + 1) = 0

Therefore , x – 2 = 0 or x + 1 =0

This implies that a=2 or a= -1

3. $y^{2}-7y+10=0$

Replace -7y with -2y -5y as follows

. $y^{2}-2y-5y+10=0$

Group your brackets:

$\left ( y^{2}-2y \right )-\left ( 5y+10 \right )=0$

Factorize your grouped terms by taking out the common factors in each group.

y( y – 2 ) – 5( y – 10 ) = 0

collect those terms outside the brackets and one of the like terms inside the brackets as shown.

( y – 5 )( y – 2 ) = 0

Then , y – 5 = 0 or y – 2 = 0

Y = 5 or y = 2

4. $m^{2}+8m+15=0$

Replace 8m with 3m+5m as follows;

$m^{2}+3m+5m+15=0$

Group your brackets:

$\left ( m^{2}+3m \right )+\left ( 5m+15 \right )=0$

Factorize your grouped terms by taking out the common factors in each group.

m(m+3) + 5(m+3)

collect those terms outside the brackets and one of the like terms inside the brackets as shown.

(m+3)(m+5)

Then , m+3 = 0 or m+5 = 0

m=-3 or m= -5

5. $4t^{2}-5t-21=0$

Replace -5t with 12t-7t as follows:

$4t^{2}-5t-21=0$

Group your brackets:

$\left ( 4t^{2}-12t \right )+\left ( 7t-21 \right )=0$

Factorize your grouped terms by taking out the common factors in each group.

4t( t + 3 ) - 7( t + 3) = 0

collect those terms outside the brackets and one of the like terms inside the brackets as shown.

( 4t - 7) ( t + 3) = 0

Then , 4t-7 = 0 or t + 3 = 0

4t = 7 or t = -3

t= 7/4 0r t= -3

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