# Algebra II

posted by on .

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 - ,

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.

(x-x1)(x-x2) = 0

x1 = (1/2a)[-b + sqrt(b^2-4ac)]
x2 = (1/2a)[-b - sqrt(b^2-4ac)]

• Algebra II - ,

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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