Question

2x+1=square root of (3x^2+5x+3) how do I solve this?

Answer 1

lmfao, that's the answer?

Answer 2

yeah

Answer 3

how'd you get it?

Answer 4

Magic :) haha

Answer 5

lol

Answer 6

It's just simplifying

Answer 7

that makes sense

Answer 8

\[(2x+1)^2=3x^2+5x+3, \text{where } 3x^2+5x+3 \ge 0\]

Answer 9

Then you simplify the quadratic, for 2 solutions of x.
Check them both against the inequality to see which, or neither, or both work. :)

Answer 10

so there can be two answers?

Answer 11

-1/3 is not the answer btw

Answer 12

okay, then once you get your two answers you check it, then what?

Answer 13

potentially if both solutions make 3x2+5x+3 greater than or equal to 0

Answer 14

An answer that statisfies both the equality and the inequality is THE answer.

Answer 15

okay, so if you check it and it works once you plug it back in, its the answer?

Answer 16

You should check BOTH solutions of the equality against the inequality.
(this example does have 2 solutions)

Answer 17

oh okay, i have another one similar to this next question. do u mind helping me with that too

Answer 18

Sure do you want me to verify your solution to this one first?

Answer 19

I already put in the wrong one. lol. I'm too trusting. but i'm going to do it correctly next

Answer 20

ok

Answer 21

the next one is x+2=square root of (-1-2x)

Answer 22

I got down to two answers -1 and -5 but i checked and -5 doesnt work. so it's -1?

Answer 23

So we have\[x+2=\sqrt{-1-2x}\] we square both sides and keep in mind that:\[-1-2x \ge 0\]
\[(x+2)^2=-1-2x\]\[x^2+4x+4=-1-2x\]\[x^2+6x+5=0\]

Answer 24

yeah, that's what i got when I simplified

Answer 25

You got it right -1

Answer 26

I think x=2 is right answer
not x=-1 check it !