[SPLIT] solve z^2 + z = 2 by factoring

[ssewell]

I am lost on Quadratic equations I have to solve this one by factoring please tell me if I have this right.

z² +z=2

(z+1) (z-1)=2

z=-1 or z=1

 

[JeffM]

I am lost on Quadratic equations I have to solve this one by factoring please tell me if I have this right.

z² +z=2

(z+1) (z-1)=2 Oh no (z + 1)(z - 1) = z(z - 1) + 1(z - 1) = z^2 - z + z - 1 = z^2 - 1 = z^2 - 1^2

z=-1 or z=1 I see where you are lost.

Please start your own thread rather than tacking on to the end od someone else's thread..

Put your quadratic equation into standard form $\displaystyle ax^2 + bx + c = 0.$ Then solve.

There are at least three ways to solve a quadratic in standard form. If it is easily factored, you get something like this

$\displaystyle (dx + e)(fx + g) = 0,\ where\ df = a,\ eg = c,\ and\ (dg + ef) = b.$

Now if 2 numbers multiply to zero, one or the other or both are equal to zero.

$\displaystyle If\ (dx + e) = 0,\ then\ x = \dfrac{-e}{d}.\ But\ if\ (fx + g) = 0,\ then\ x = \dfrac{-g}{f}.$

Put your equation into standard form and then try to factor. Now what?

 

[HallsofIvy]

I am lost on Quadratic equations I have to solve this one by factoring please tell me if I have this right.

z² +z=2

(z+1) (z-1)=2

z=-1 or z=1

Have you never actually taken an algebra class? First, $\displaystyle z^2+ z$ does NOT factor as (z- 1)(z+ 1). Multiplying those gives $\displaystyle z^2- 1$. $\displaystyle z^2+ z$, because there is at least one "z" in each term, factors as z(z+ 1).

But factoring does not help here. The fact that "ab= 2" does NOT mean that "a= 0 or b= 0". That is a property only of 0: if ab= 0 then either a= 0 or b= 0. You need to first get it equal to 0: $\displaystyle z^2+ z- 2= 0$. Now, factor $\displaystyle z^2+ z- 2$. That should be easy because -2 factors only as (2)(-1) or (-2)(1).

Finally, do you even understand what "solve an equation" means? z= -1 is NOT a solution because $\displaystyle (-1)^2+ (-1)= 1- 1= 0$, NOT 2.