Question

would (x+1)(x+1)=4x^2 ?

Answer 1

2x+2

Answer 2

so no

Answer 3

how come?

Answer 4

no
it would equal \[x^2 +x +x +1 = x^2 +2x+1\]

Answer 5

(x+1)(x+1) = x^+2x +1

Answer 6

you add your two like terms, which equals 2x and then add your constants(which is 2)

Answer 7

what r the 2 like terms?

Answer 8

never mind i forgot to multiply but like terms r (x) bcuz they both end in (
x) so 4x is a like term with x but x^2 and 4x r not like terms

Answer 9

oh, ok

Answer 10

\[x^2+2x+1=4x^2\]
\[0=4x^2-x^2-2x-1\]
\[0=3x^2-2x-1\]
\[0=3x^2-3x+x-1\]
\[0=3x(x-1)+1(x-1)\]
\[0=(x-1)(3x+1)=> x=1 \text{ or } x=\frac{-1}{3}\]
so
\[(x+1)(x+1)=4x^2 \text{ when } x=1 \text{ or } x=-\frac{1}{3}\]

Answer 11

@myininaya, what does that have 2 do with my problem?

Answer 12

its when those two expressions are equal

Answer 13

what 2 expressions?

Answer 14

4x^2
and
(x+1)(x+1)

Answer 15

@myininaya thts what i said but u wrote it out and did somesort of mathy fun stuff

Answer 16

lol

Answer 17

how r they equal?

Answer 18

they are only equal when x=1 or x=-1/3
they are not equal for any other numbers

Answer 19

why is that?

Answer 20

4x^2 and (x+1)(x+1) do not intersect anywhere else

Answer 21

they have the same variable

Answer 22

@myininaya. what do u mean by intersceting any where else?

Answer 23

their graphs only meet at x=1 or x=-1/3

Answer 24

you can give a counterexample of why (x+1)(x+1)=4x^2 do not equal for all x
if x=0, (x+1)(x+1)=(0+1)(0+1)=1(1)=1
4x^2=4(0)^2=4(0)=0
1 does not equal 0

Answer 25

i'm confused at where u r heading

Answer 26

you asked why they only equal for two values of x
when we moved everything to one side we had a 2nd degree polynomial
which means there are two solutions
and both of those solutions were real (we got two different real solutions)
so their graphs only interest in two different spots which were x=1 or x=-1/3 as showed above

Answer 27

intersect*

Answer 28

when u r talking about graphs and interseceting and stuff, what does that mean?

Answer 29

you haven't heard of a graph?

Answer 30

well, i have when it comes 2 comparing stuff and i did something similar 2 that in math class a week ago, so no not really