I am having problems figuring out how to solve the following second degree equations:
- 2x$^2$ + 3x + 1 = 0
I can't get factors that add together to get 3 or multiply together to get 1:
(2x + ?)(x + ?)
- 4x$^2$ + x = 0
I am not sure how to factorise these.
- 0 = 4y$^2$ + 8y
I can factor this to 4[(y + 1)(y + 1)] but I'm not sure what to do after that.
- x$^2$ = -3x
I I got x = -3 for this but the answer is x = 0 or x = -3, I don't get where the 0 came from.
I have the answers so it is really the explanation that is important to me.
For case 1: realize that $(2x+a)(x+b)=2x^2+(a+2b)x+ab$ and start searching for $a,b$ such that $ab=1$ and $a+2b=3$ (so not $a+b=3$ as you seem to think).
For case 2: realize that $x$ is a common factor of $4x^2$ and $x$ so that you can write $4x^2+x=x(4x+1)$
A common factor can also be found in case 3. Your factorization is incorrect.
Concerning case 4: always start with "reduction to zero". That is: realize that $x^2=-3x$ is the same statement as $x^2+3x=0$. Again a common factor can be found.