Please use factoring to solve for x in the following question:

Question

12X^2 - 72X = 108
63X^2 - 112 = 0

Answer 1

 

12 X² - 72X -108 = 0

 

first factor 12:

 

12( X² - 6X + 9)  =0

 

    Now to factor quadratic,  think of 2 numbers whose sum is 6 and their product 9. Have to do this in your head or scratch paper, or .... the numbers are 3, 3 , 3*3 = 9 , 3+3 =6, then the quadratic factors to:

 

    12 ( X -3 ) ²  = 0     x=3 , is repeated roots of the quadratic

 

     63 X² - 112 =0

   

       7 ( 9X² - 16 ) =0

 

          9 X² - 16 =0

 

           X² = 16/9          X =±4/3

 

 ( X + a) ( X + b) = X²  + ( a +b) X + ab (1)   ( X + a) ² = X² + 2aX +a^{2 (2)}   ( X +a ) ( X -a )= X² -a^{2 (3)} 

 

 The 3 above identities are key to factoring quadratic, and finding the roots

 

    given a quadratic  like :  X²  + 7X + 10

 

 here we see that ( a + b) = 7   ab =10

   a = 2 b =5 is the answer, therefore  X²   + 7X + 10 = ( X +2) ( X + 5 )

   equation (2) is a special case of (1) where a=b , a+b = 2a , ab=a²

   equation (3) is a special case of ( 1) where b = -a, a + ( -a) = 0 , ab = a²

 

  These 3 identities are used in factoring a quadratic.

   

    Equation (1) is factorable if there exists 2 whole number whose sum is ( a+ b), and product =ab.

 

     If the answer of the system of equation is not a whole number, then have to do factoring by competing the square, and come up with a factors of irrational and complex numbers, yielding to Irrational and complex roots . 

  

 

Answer 2

The second one is easy to factor.

 

63x²-112=0 can be factored using property a²-b²=(a-b)*(a+b)

 

63x²-112=7*(9x²-16)=7*(3x-4)*(3x+4). I took factor 7 out to make it look not ugly.

 

So, the equation becomes 7*(3x-4)*(3x+4)=0, from which it follows that x=4/3 or x=-4/3.

 

In the first case,

factor out 12 first.

 

12x²-72x=108 ⇔ 12*(x²-6x)=108 ⇔ x²-6x=9

 

Now move 9 to the left to get:

 

x²-6x-9=0; Now this can be factored as (x-...)*(x-...), the factors in both parentheses having to multiply to -9 and adding to 6. It is not quite obvious how to choose those factors. This is why I despise the factoring as the main way of solving quadratic equations!

 

Here I suggest to add and subtract 9 to the left part. You will get:

 

x²-6x+9-18=0 or x²-6x+9=18

 

Now the expression on the left side can be easily factored, since two factors now have to sum to 6 and multiply to 9. Those are 3 and 3. Thus we get:

 

(x-3)²=18

From here you get:

x-3=±3√2

So,

x₁=3+3√2

x₂=3-3√2