Question

4x =√(2x + 3)

Answer 1

square both sides, start with \(16x^2=2x+3\)

Answer 2

Am I going to get two answers then?

Answer 3

probably.

Answer 4

I should only get 1 though.

Answer 5

because you have a quadratic, correct?

Answer 6

I'm honestly not sure.

Answer 7

\[16x^2 -2x -3 = 0\]\[x ={ 2\pm \sqrt{(-2)^2 -4(16 \times (-3))} \over {2 \times 16}}\]

Answer 8

After you square both sides, you get the form Ax^2 + Bx + c, which you should recognize as a quadratic, giving 2 possible answers.

Answer 9

right, but I shouldn't get two answers because I'm inputting these on an online site that only allows one answer.

Answer 10

what are the options?

Answer 11

you need so check your answers in the original equation since step one was to square

Answer 12

It's not giving any options unfortunately.

Answer 13

try putting in 1 of the answers.

Answer 14

I tried & neither worked. I had gotten 1/2 & 3/8

Answer 15

\[16x^2-2x-3=0\]
\[(2 x-1) (8 x+3) = 0\]
\[x=\frac{1}{2},x=-\frac{3}{8}\] now check each one

Answer 16

None work, according to the online thing at least.

Answer 17

\[x=\frac{1}{2}\]
\[4\times \frac{1}{2}=\sqrt{2\times\frac{1}{2}+3}\]
\[2=\sqrt{1+3}=\sqrt{4}=2\] is true, so \(x=\frac{1}{2}\) is a solution

Answer 18

Is the question asking for x?

Answer 19

For whatever reason I plugged in .5 in the answer key thing & it worked. Though 1/2 didn't.

Answer 20

\(-\frac{3}{8}\) will not work because the left hand side will be negatve and the right hand side will be positive

Answer 21

Okay thanks guys :D It works.

Answer 22

on line class! no wonder

Answer 23

Yeah it's not cooperative. I keep putting the questions on here because after plugging a bunch in, it usually works.