That equation is same as: x^2 + 6x - 27 = 0nae said:2. x^2-3x+9x=27
It can be solved by using the Quadratic Formula, right? It simplies, as you know, to 7x^2-4x+0=0 which as long as a is not equal to zero then it is possible to solve, Right?stapel said:The first one is already in factored form. .
nae said:2. x^2-3x+9x=27
I am lost.
No need for quadratic, jonboy:jonboy said:It can be solved by using the Quadratic Formula, right? It simplies, as you know, to 7x^2-4x+0=0 which as long as a is not equal to zero then it is possible to solve, Right?stapel said:The first one is already in factored form. .
mcrae said:jonboy, you keep saying that binomials cannot be equal to 0 when using the quadratic formula, then list them as ax^2 + bx + c = 0
this was your only post where you mentioned that it is actually a that cannot be 0
just watch out for little errors like that that could completely confuse someone
jonboy said:Since you can put \(\displaystyle \bold p^2+6p+4=0\) in the form \(\displaystyle ax^2 + bx + c = 0\) in which a is not=0, yes it can be factored by using the Quadratic Formula.
jonboy said:As long as \(\displaystyle a\) is not equal to zero and can be expressed using \(\displaystyle ax^2+bx+c=0\) then it can be solved by using the Quadratic Formula.