Question

Both solutions are extraneous because they don't satisfy the original equation.

What error did Eloise make?

Answer 1

here is the question:

Answer 2

any thoughts ?

Answer 3

i tried solving on my own but i got stuck and this part:

\[(\sqrt{-2x})^2 = (x+2)^2\]

Answer 4

how would i square x? do i square both altogether?

Answer 5

im talking about the (x+2)^2 part

Answer 6

There is a mistake before that, so let's not look at that.

Instead, look at
\[ \sqrt{-2x+1}-1 \]
remember, you can think of square roots as exponents, like this:
\[ (-2x+1)^\frac{1}{2} -1 \]

you can't subtract off the 1 inside the parens.

Answer 7

if you have a calculator
you will find
\[ \sqrt{5}-1 \ne \sqrt{4} \]

Answer 8

the last good step is
\[ \sqrt{-2x+1}= (x+3) \]
and what we do next is square both sides
what do we get on the left side after we square it ?

Answer 9

squaring a square root "gets rid" of the square root sign

Answer 10

why can't i subtract the one? and what did you do to put in the calculator to get that and how does that tell you it's incorrect to subtract the one? how would i know if what i subtract is the right thing to do or not? (a lot of questions sorry, i just want to learn the subject well)

on the left side i got: -2x +1

i couldn't solve the right side

Answer 11

Did you learn order of operations ?

Answer 12

yes! PEMDAS right?

Answer 13

ok, so you know (for example)
2*2 - 1
you do 2*2 first to get 4
and then 4-1
does that make sense ?

Answer 14

and you do *not* do the 2-1 first:
2*2-1 becomes 2* 1 = 2 <-- wrong.

Answer 15

yes that makes sense!

Answer 16

The point is, you are only allowed to add or subtract after you first do parens, exponents and multiply or divide. (pretty low on the totem pole)
square roots (and *everything* inside them) are done first
so for example
\[ \sqrt{4} -1 \]
is not \[ \sqrt{4-1} \]
those are two different things

Answer 17

I hope you don't have to solve Eloise's problem because it requires using the quadratic formula, and if you don't know how to do (x+3)^2 you probably don't know the quadratic formula ... though perhaps it's in your notes?

Answer 18

oh i think i understand, to subtract from the one you would just have to square it first to get: -2x +1

i don't have to solve her problem but i want to anyway, you don't have to help me with the rest of it if you don't want. but thank you thus far!

Answer 19

This shows how to multiply (x+3)(x+3)
https://www.khanacademy.org/math/algebra-basics/quadratics-polynomials-topic/multiplying-binomials-core-algebra/v/multiplying-binomials

Answer 20

and this one
https://www.khanacademy.org/math/algebra-basics/quadratics-polynomials-topic/multiplying-binomials-core-algebra/v/special-products-of-polynomials-1

Answer 21

so far you have
\[ \sqrt{-2x+1}= (x+3) \]
square both sides:
\[ -2x+1= (x+3)^2 \]
remember, (x+3)^2 means (x+3)(x+3)
if we multiply that out (see videos) we get \[ x^2+6x+9\]
so the problem is
\[ -2x +1 = x^2+6x+9\]
next add +2x to both sides
\[ 2x -2x + 1 = x^2 +6x+2x+9\]
on the left side 2x's take away 2x's is 0
on the right side, 6 x's plus 2 more x's is 8 x's
we get
\[ 1 = x^2+8x+9\]

Answer 22

next, add -1 to both sides
\[ 1+-1 = x^2+8x+9+-1\]
on the left side we get 1-1=0
on the right, simplify 9-1 to get 8
we have
\[ 0 = x^2+8x+8\]
which is usually written this way:
\[ x^2+8x+8=0\]

Answer 23

that equation is solved using the quadratic formula
https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/v/using-the-quadratic-formula
which takes a bit of time to understand and use.

Answer 24

the two answers you get are
\[ -4 + 2\sqrt{2} \]
and \[ -4 -2 \sqrt{2} \]
the 2nd solution is *extraneous*
and we need a calculator to evaluate these messy numbers.

Answer 25

oh okay, i think i remember a bit of the formula! could you not simplify those expressions into \[-2\sqrt{2}\] and \[-6\sqrt{2}\]?

Answer 26

no, the \[ 2 \sqrt{2} \]
means 2 times sqr(2) i.e. multiplying two numbers
in other words that is like doing the -4+2 first in
-4 + 2*2
to get -2*2 = -4
but we are not allowed to do that

Answer 27

remember add/subtract is very low priority, and multiply comes first

Answer 28

oh right! due to pemdas, right!

Answer 29

if you see two numbers being multiplied, you can't subtract or add anything to either of them.
1+2*3
if you added the 1 to the 2 first, we would get 3*3
but we know we have to do 2*3 first and the answer is 1+6=7

Answer 30

so try to think like this:
square undoes square root
divide undoes multiply
subtract undoes add (and vice versa)

Answer 31

oh this is making so much sense to me thank you so much! i'm gonna try to solve the rest of the problem using the two answers, but thank you for helping me understand the ins and outs because I overlook them so often and make "silly mistakes"! thank you again