Hi!
Could someone please solve this equation for me and try to explain how you did that. 4*(x-1)^2+1/3=7/9
I do know the answer for this equation. But I don't know how to think when solving it and how to get the answer.
Hi!
Could someone please solve this equation for me and try to explain how you did that. 4*(x-1)^2+1/3=7/9
I do know the answer for this equation. But I don't know how to think when solving it and how to get the answer.
$\displaystyle 4(x-1)^2 + \frac{1}{3} = \frac{7}{9}$
multiply every term by 9 to clear the fractions ...
$\displaystyle 36(x-1)^2 + 3 = 7$
subtract 3 from both sides ...
$\displaystyle 36(x-1)^2 = 4$
divide both sides by 36 and reduce ...
$\displaystyle (x-1)^2 = \frac{1}{9}$
"un"square both sides ...
$\displaystyle x-1 = \pm \frac{1}{3}$
add 1 to both sides ...
$\displaystyle x = 1 \pm \frac{1}{3}$
so, the two solutions are $\displaystyle x = \frac{4}{3}$ and $\displaystyle x = \frac{2}{3}$
Thank you very much skeeter!
Never realised that it was so easy.
HallsofIvy: Well, not really :P I just translated the name we use for this kind of equations here in Sweden directly to english. So It maybe isn't very correct
check this out ...
http://http://www.wired.com/magazine/2011/07/ff_khan/all/1
... won't be long until we're all out of a job.
Okey, here you've got one more equation to help me out with. I have to say that this book really sucks at explaining (and so does my math teacher also, sadly enough).
2x^2+4x+40=3x^2-2x
And by the way. It would be extreamly nice if you could help me out two factorizing problems also:
1. 2x+2a2b(x+a)
2. a^(x+2)-a^(2x)
Put all the terms to one side, so you get:
$\displaystyle 2x^2+4x+40-3x^2+2x=0$
$\displaystyle \Leftrightarrow -x^2+6x+40=0$
Use the quadratic formula to continue.
Ex 2:
Determine which factor the terms have in common, for example:
2. $\displaystyle a^{x+2}-a^{2x}=a^{x}\cdot a^{2} - (a^{x})^2=...$