Posted by Anonymous on Tuesday, May 18, 2010 at 6:21pm.
For the : ax^2 + bx + c
How would you get what to put in
(_+_)(_+_)= 0
Is it c doubled and then finding the factors that could equal b or 1/2 of b times c...?
Could someone post how I could find the answer this way (shown above)
(AND NOT THE QUADRATIC FORMULA)?
Thnak You.

Algebra II  drwls, Tuesday, May 18, 2010 at 11:13pm
ax^2 + bx + c
is NOT an equation.
Why should you expect to be able to rewrite it as (_+_)(_+_)= 0 ?
The values of x that satisfy the equation
ax^2 + bx + c = 0
ARE given by the quadratic formula.
Call them x1 and x2.
(xx1)(xx2) = 0
x1 = (1/2a)[b + sqrt(b^24ac)]
x2 = (1/2a)[b  sqrt(b^24ac)]

Algebra II  Anonymous, Tuesday, June 15, 2010 at 11:07pm
a ( x^2 + (b/a) x + (c/a))
a ( x^2 + 2(b/(2a)) x + (c/a))
a [ x^2 + 2(b/(2a)) x + (b/(2a))^2 +
 ( (b/(2a))^2  (c/a) )]
a [( x + (b/(2a)))^2 
 ( (b/(2a))^2  (c/a) )]
a [( x + (b/(2a)))^2 
 ( (b^2  4ac/((2a)^2)]
a [( x + (b/(2a)))^2 
 ( (b^2  4ac/((2a)^2)]
a [( x + (b/(2a)))  ( sqrt(b^2  4ac/((2a))] *
[( x + (b/(2a))) + ( sqrt(b^2  4ac/((2a))]
a [( x + ((b/(2a))  sqrt(b^2  4ac/((2a))] *
[( x + (b/(2a))) + sqrt(b^2  4ac/((2a))]
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