Solve Quadratic Equation: y^2 - 12y + 32 = 0

[fbsthreads]

in a maths book i have there is an example of solving a quadratic. part of the process reads:

y^2 - 12y + 32 = 0


implies that


(y-8) (y-4) = 0



i don't understand how this second equation was reached based on the first one.

could someone add in the steps that gets me from the first equation to the implication.

thanks.

 

[jcsd]

in a maths book i have there is an example of solving a quadratic. part of the process reads:

y^2 - 12y + 32 = 0


implies that


(y-8) (y-4) = 0



i don't understand how this second equation was reached based on the first one.

could someone add in the steps that gets me from the first equation to the implication.

thanks.

expand:

(y + a)(y + b)

and you get:

y^2 + (a+b)y + ab

so from your equation above you know that:

a + b = -12

and

ab = 32

as

-8 + -4 = -12

and

-8*-4 = 32

a and b must be -8 and -4.

 

[xt]

i don't think you can, it is obvious.

if you actually want the logics of thinking its like this:

we want to express the quadratic as (y-a)(y-b),

now a*b = 32

and a + b = 12

so try all the integers you can think of, you'll find 8 and 4 does.

 

[arildno]

A simple way uses the idea of "completing the square:
$$y^{2}-12y+32=0$$
Now, regarding -12=2*(-6), we add 0 to our equation in this manner:
$$y^{2}-12y+32+(-6)^{2}-(-6)^{2}=0$$
Or, equivalently, for the Left Hand Side:
$$y^{2}-12y+32+(-6)^{2}-(-6)^{2}=(y^{2}-2*6y+6^{2})+(32-6^{2})=(y-6)^{2}-4$$
Furthermore, since $$4=2^{2}$$ we have:
$$(y-6)^{2}-4=(y-6)^{2}-2^{2}=(y-6+2)(y-6-2)=(y-4)(y-8)$$

Finally, by setting this expression (which is equivalent to our original left hand side) equal to our originil right hand side (that is,0) we gain:
$$(y-4)(y-8)=0$$
as required.

 

[fbsthreads]

thanks people.
it was the bit i have highlighted in bold that i had forgotten about when trying to work it out.

y^2 + (a+b)y + ab

thanks again.