I am having trouble with two problems. 1. Edit: figured it out 2. 3(x+7/3)^2 = 1 I'm just lost on that one D: Thanks for any help you guys have for me
Follow Math Help Forum on Facebook and Google+
Originally Posted by ajm490 I am having trouble with two problems. 1. (x-3/5)^2 = 4/5 I turn it into x^2 -6/5X+9/25= 4/5 and then x^2 -6/5X-11/25 = 0. Then I get stuck trying to plug that into the quadratic formula. $\displaystyle (x-3/5)^2 = 4/5 $ so: $\displaystyle x-3/5 = \pm \sqrt{4/5} $ hence: $\displaystyle x = 3/5 \pm \sqrt{4/5} $ No need for the quadratic formula RonL
Originally Posted by ajm490 2. 3(x+7/3)^2 = 1 do not delete your questions after they have been solved $\displaystyle 3(x + 7/3)^2 - 1 = 0$ $\displaystyle \Rightarrow 3 \left( x^2 + \frac {14}3x + \frac {49}{9} \right) - 1 = 0$ can you take it from here?
Yes, I think so. And I found the answer by myself actually, prior to him completing it, which is why I took it off. Thanks to both of you though.
Originally Posted by ajm490 I am having trouble with two problems. 1. Edit: figured it out 2. 3(x+7/3)^2 = 1 I'm just lost on that one D: Thanks for any help you guys have for me Same as other one: $\displaystyle 3(x+7/3)^2 = 1 $ so: $\displaystyle (x+7/3)^2 = 1/3 $ then: $\displaystyle x+7/3 = \pm \sqrt{1/3} $ hence: $\displaystyle x = -7/8 \pm \sqrt{1/3} $ RonL
Originally Posted by CaptainBlack Same as other one: $\displaystyle 3(x+7/3)^2 = 1 $ so: $\displaystyle (x+7/3)^2 = 1/3 $ then: $\displaystyle x+7/3 = \pm \sqrt{1/3} $ hence: $\displaystyle x = -7/8 \pm \sqrt{1/3} $ RonL i was working too hard
Originally Posted by Jhevon i was working too hard I had thought of mntioning that, but then .. RonL
View Tag Cloud