Question

\(x^2-2x+3=x^3+2x+x^2\) is equivalent to

\(0\)
\(2x^2-4x=0\)
\(-x^3+4x-3=0\)
\(x^3-2x^2-3=0\)
\(x^3+4x-3=0\)

Answer 1

YOur idea?

Answer 2

subtract x^2 , -2x and 3 in turn from the left hand side and see what you get

Answer 3

Okay. Hold on..

Answer 4

what do you think ?

Answer 5

I really dont know.

Answer 6

\[\large \bf x^2-2x+3=x^3+2x+x^2\]
first subtract x^2 to both sides,

Answer 7

what we get ?

Answer 8

\(x^2\) cancels out..

Answer 9

Equivalent to 0

Answer 10

How? Show me..

Answer 11

when x^2 cancels out , we get
\[\large \bf -2x+3=x^3+2x\]
Add 2x to both the sides

Answer 12

what we get ?

Answer 13

Use, http://mathpapa.com/algebra-calculator.html

Answer 14

We get..
\(-2x+2x+3=x^3\)

Answer 15

nope !
try again

Answer 16

But I added the \(2x\) from the other side..

Answer 17

Oh.. to both sides.
@mayankdevnani

Answer 18

I got \(4x\) on the right side.

Answer 19

so we get
\[\large \bf 3=x^3+4x\]
now subtract 3 to both sides,we get
\[\large \bf x^3+4x-3=0=\color{red}{Answer}\]

Answer 20

hope you understand ! @TheRaggedyDoctor

Answer 21

Oh, okay. Thank you. Yes! I understood.

Answer 22

your welcome :)