y= 3(2x + 1)^2 - 5. Please show all working out >.<

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- Apr 13th 2008, 03:03 AMdelicate_tearsSolve for x- REALLY REALLY URGENT
y= 3(2x + 1)^2 - 5. Please show all working out >.<

- Apr 13th 2008, 03:05 AMMoo
Hello,

Why don't you try a little on your own ?

And what exactly d'you want ? Solve for y=0 ? Or express x with y ? - Apr 13th 2008, 03:22 AMdelicate_tears

I wanted to solve for x, hence the topic. I tried it before , but I got (-1 + square root 15) /2, which was totally not the answer. That's why I wanted the steps, to show me how to get there.*I know the answer, it's 0.145... and -1.145... I just need to know how to get round it.* - Apr 13th 2008, 03:26 AMjanvdl
$\displaystyle y = 3(2x + 1)^2 - 5$

$\displaystyle y + 5 = 3(2x + 1)^2$

$\displaystyle \frac{y + 5}{3} = (2x + 1)^2$

$\displaystyle \sqrt{ \frac{y + 5}{3} } = 2x + 1$

$\displaystyle \sqrt{ \frac{y + 5}{3} } - 1 = 2x$

$\displaystyle \frac{ \sqrt{ \frac{y + 5}{3} } - 1 }{2} = x$ - Apr 13th 2008, 03:28 AMMoo
*You want to solve for x in 3(2x + 1)^2 - 5=0*

This is completely different than your previous question ^^

Well

3(2x+1)²=5

(2x+1)²=5/3

2x+1=sqrt(5/3) or -sqrt(5/3) remember this ! (a)²=(-a)²=a² (Wink)

2x=-1+sqrt(5/3) or -1-sqrt(5/3)

And then divide by 2. - Apr 13th 2008, 03:35 AMMoo