12X^2 - 72X = 108

63X^2 - 112 = 0

63X^2 - 112 = 0

12X^2 - 72X = 108

63X^2 - 112 = 0

63X^2 - 112 = 0

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Woodland Hills, CA

12 X^{2} - 72X -108 = 0

first factor 12:

12( X^{2} - 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 X^{2} - 112 =0

7 ( 9X^{2 }- 16 ) =0

9 X^{2} - 16 =0

X^{2} = 16/9 X =±4/3

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

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

given a quadratic like : X^{2 }+ 7X + 10

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

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

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

equation (3) is a special case of ( 1) where b = -a, a + ( -a) = 0 , ab = a^{2}

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 .

Willowbrook, IL

The second one is easy to factor.

63x^{2}-112=0 can be factored using property a^{2}-b^{2}=(a-b)*(a+b)

63x^{2}-112=7*(9x^{2}-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.

12x^{2}-72x=108 ⇔ 12*(x^{2}-6x)=108 ⇔ x^{2}-6x=9

Now move 9 to the left to get:

x^{2}-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:

x^{2}-6x+9-18=0 or x^{2}-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,

x_{1}=3+3√2

x_{2}=3-3√2

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