1. ## Solving Quadratic Equations/ factoring

Please help, I'm in trig and we are reviewing algebra, having a hard time remembering how to do some of this stuff. Thank you!

Factor Completely if possible:

8.) me + mi + my

Solve each equation for x:

48.) (2x-3)(2x-3) = 4

52.) a.) 12x^2 - 5x -3 = 0
b.) 30x^2 + 13x -10 = 0

2. Originally Posted by iDrum

Factor Completely if possible:

8.) me + mi + my
Take out $\displaystyle m$ as a common factor

Spoiler:
$\displaystyle m(e + i + y)$

Originally Posted by iDrum

Solve each equation for x:

48.) (2x-3)(2x-3) = 4

52.) a.) 12x^2 - 5x -3 = 0
b.) 30x^2 + 13x -10 = 0

Exapnd the first equation to find the form $\displaystyle ax^2+bx+c=0$

Then for that and the other 2 use

$\displaystyle x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

3. Hello iDrum
Originally Posted by iDrum
...
Solve each equation for x:

48.) (2x-3)(2x-3) = 4
Write this as:
$\displaystyle (2x-3)^2=4$
and then take square roots, not forgetting the $\displaystyle \pm$ sign on the right:
$\displaystyle 2x-3=\pm2$

$\displaystyle \Rightarrow 2x = 3+2$ or $\displaystyle 3-2$

$\displaystyle \Rightarrow x = 2.5$ or $\displaystyle 0.5$
52.) a.) 12x^2 - 5x -3 = 0
$\displaystyle 12x^2 - 5x -3 = 0$

$\displaystyle \Rightarrow(3x+1)(4x-3)=0$

Can you complete it?
b.) 30x^2 + 13x -10 = 0
This won't factorise, so use the formula:
$\displaystyle x=\frac{-13\pm\sqrt{13^2-4(30)(-10)}}{60}$
Can you complete it?