Changing RHS to 0

[Guest]

Hi! i'm just curious, why must we equate the equation to 0 when solving a quadratic equation? For example, changing [tex]2x^2-3x=-1[/tex] into [tex]2x^2-3x+1=0[/tex]. Thankss. Sorry if this is a really basic question sometime I just don't get how stuff works-

[HallsofIvy]

I hope this won't surprise you! Since there is a "-1" on the right side, you get 0 by adding 1! Of course whatever you add on the right side you must add on the left side.

$2x^2- 3x= -1$
$2x^2- 3x+ 1= -1+ 1$
$2x^2- 3x+ 1= 0$

[Guest]

The reason we do that is the "factor property": if ab= 0 then either a= 0 or b= 0 or both.
[tex]2x^2- 3x+ 1= (2x- 1)(x- 1)= 0[/tex].
Either 2x- 1= 0 so x= 1/2 or x- 1= 0 so x= 1.

But we don't have to do it and don't always do it. Many quadratic expressions do not factor so easily but we can always "complete the square": [tex]2x^2- 3x= 2(x^2- (3/2)x)= 2(x^2- (3/2)x+ 9/16- 9/16)= 2(x^2- (3/2)x+ 9/16)- 9/8= 2(x- 3/4)^2- 9/8= -1[/tex], [tex]2(x- 3/4)^2= -1+ 9/8= 1/8[/tex]. So [tex](x- 3/4)^2= 1/16[/tex], [tex]x- 3/4= \pm 1/4[tex]. [tex]x= 3/4+ 1/4= 1[/tex] or [tex]x= 3/4- 1/4= 1/2[/tex].