Question

Put the equation below in the form ax2 + bx + c = 0. Enter exponents using the caret ( ^ ). For example, you would enter 4x2 as 4x^2.

-x2 + 3 = (2x - 1)2

Answer 1

Ok, right now, your equation is:
-x2 + 3 = (2x - 1)2

Let us insert the caret sign first. Do this were there needs to be a square root
Do you know where?

Answer 2

yush i think.
-x^2+3=(2x-1)^2

Answer 3

Yes, good job. Now its:
-x^2+3=(2x-1)^2

To make it more visibaly appealing, I will make it look a bit more fancy
\[-x^2+3=(2x-1)^2\]
Now, we know that when we square a number, we mean multiply that number by itself 2 times. So we can rewrite that as:
\[-x^2+3=(2x-1)(2x-1)\]
Expand the brackets

Answer 4

Try it @jackelyn_1234

Answer 5

expand brackets?

Answer 6

Yes. See the:
\((2x-1)(2x-1)\)
Expand them

Answer 7

Use the foil method

Answer 8

ohh! first outer inner last

Answer 9

Yes, EXACTLY :)
Do that for the brackets

Answer 10

4x^2-4x+1

Answer 11

Yes, good. Now we have:
\[-x^2+3=4x^2-4x+1\]
Now, we wan't to move everything so all the stuff is on one side. Can you do that?
Remember, when everything is moved from one side, all we have left is 0

Example:
\(x+2=2\)
\(x+2-2=0\)

Answer 12

want*

Answer 13

-5x^4+4x+4
i believe

Answer 14

I don't believe so. Here, let me help you.
Equation:
\[-x^2+3=4x^2-4x+1\]
To move the \(3\) to the right, we need to minus \(3\) to both sides, since the \(3\) is a positive number. \[-x^2+3-3=4x^2-4x+1-3\]Adding three and taking away 3 cancel each other on the left hand side. \[-x^2\cancel{+3-3}=4x^2-4x+1-3\]\[-x^2=4x^2-4x+1-3\]
Do the same thing with the \(-x^2\). This time, add \(-x^2\), since it is a negative number
\[-x^2=4x^2-4x+1-3\]\[-x^2+x^2=4x^2-4x+1-3+x^2\]\[\cancel{-x^2+x^2}=4x^2-4x+1-3+x^2\]\[0=4x^2-4x+1-3+x^2\]\[4x^2-4x+1-3+x^2=0\]
Now simplify the left hand side.
\[4x^2-4x+1-3+x^2=0\]\[5x^2-4x+1-3=0\]\[5x^2-4x-2=0\]

Answer 15

Does that make sense, @jackelyn_1234?

Answer 16

Hello, @jackelyn_1234?

Answer 17

i apologize. my internet wifi is acting up and my internet goes off and on

Answer 18

and yes it does. i see my mistake

Answer 19

Ok, good :)

Answer 20

We did it. It is now in general form \((ax^2+bx+c)\)
\(5x^2−4x−2=0\)
In less pretty writing: 5x^2−4x−2=0

Answer 21

Hope that helped you, @jackelyn_1234 :)

Answer 22

thank you :D

Answer 23

No problem @jackelyn_1234. But man, did it take a LOT of typing to get the equations to look pretty ;)
\[\begin{bmatrix}
\color{red}{WORTH}& \color{blue}{IT!}
\end{bmatrix}\]