Solve for x

**delicate_tears** · April 13th 2008, 03:03 AM

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

**Moo** · April 13th 2008, 03:05 AM

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 ?

**delicate_tears** · April 13th 2008, 03:22 AM

Originally Posted by Moo

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 ?

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.

**janvdl** · April 13th 2008, 03:26 AM

Originally Posted by delicate_tears

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

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

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

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

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

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

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

**Moo** · April 13th 2008, 03:28 AM

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

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

And then divide by 2.

**Moo** · April 13th 2008, 03:35 AM

Originally Posted by janvdl

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

OR

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

Don't forget it

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