Question

lincoln and gemma are looking at the equation sqrt (2x + 3) = x. Lincoln says that the solution is extraneous. Gemma says the solution is non extraneous Whos correct? MEDAL WILL BE GIVEN

Answer 1

@Becki

Answer 2

@agent0smith

Answer 3

Square both sides and set it all equal to zero. Factor. Then solve. You'll get 2 answers. Put each one in for x in the original equation. Are either of your equations with each of your 2 answers true? If yes, then nonextraneous. If no, then extraneous. :)

Answer 4

wait so 2x + 3 = 0 and x = 0?

Answer 5

@Becki

Answer 6

No. [sqrt (2x+3)]^2=x^2 So if you square the square root of something you get...?? :) Think of (sqrt2)*(sqrt2)

Answer 7

(2x + 3)(2x+3)

Answer 8

Not exactly. It would be sqrt(2x+3)sqrt(2x+3) which would just equal (2x+3).

So on one side you have (2x+3) and on the other x^2.

\[2x+3=x ^{2}\] Now put everything on the same side, setting it equal to 0.

\[x ^{2}-2x-3=0\] and factor (x- ?)(x+ ?)=0

Answer 9

ehh...im lostt

Answer 10

why is it not +3

Answer 11

Because we moved everything to the side with the x^2.

2x+3=x^2
2x =x^2-3
0 =x^2-2x-3

I just turned it around so the zero was on the right.

Answer 12

2x+3=x^2
2x+3-3=x^2-3 (subtract 3 from both sides)
2x-2x=x^2-2x-3 (subtract 2x from both sides)

Answer 13

do u think u can do it with the equation button below!

Answer 14

this 2x + 3 = x^2 is confusing now. Where did u get that

Answer 15

\[2x+3-3=x ^{2}-3\]
\[2x-2x=x ^{2}-2x-3\]
\[0=x ^{2}-2x-3x \]
Turned around:
\[x ^{2}-2x-3=0\]

Answer 16

2x+3=x^2 came from squaring both sides.

If you take \[\sqrt{2x+3}\times \sqrt{2x+3}\] you get \[2x+3\]

Answer 17

ok got that part !

Answer 18

@Hero

Answer 19

I say lincon is correct

Answer 20

I just need the steps on how to cause they want me to type it

Answer 21

ah.... then nvm haha

Answer 22

id go with hero on this

Answer 23

@Luigi0210 @galacticwavesXX @recon14193

Answer 24

\[\Large x ^{2}-2x-3=0\] do you understand up to here?

Answer 25

Yes!

Answer 26

K... can you factor it? you want two numbers that multiply to make the -3, and add up to -2

Answer 27

I cant remember how to factor to well. which two numebers do u use

Answer 28

what numbers multiply to make -3? list them (there are not many)

Answer 29

1 and -3 .. and idk I only came up with one

Answer 30

when you have \[x^2-2x-3=0\] this polynomial can be factored or simplified

\[x^2-2x-3=0 \rightarrow (x-3)(x+1)=0\]

-3 + 1 = -2 that is where you get the -2x in the polynomial and -3*1 = -3 that is where the -3 came from x*x =x^2 and putting it together gives you the polynomial

you understand that so far?

Answer 31

yess!

Answer 32

squaring both sides is the standard way of solving this kind of problem

Answer 33

what happened to the sqrt in your solving method

Answer 34

haha it's all good

Answer 35

@crystelle x=3 and x=-1
because (x-3)=0 & (x+1)=0, you solve for x

Answer 36

now that you know what the solutions for x are plug in x into the original equation to check if the solutions are extraneous or not

\[\sqrt{2x+3}=x\] plug in x = 3 and x =-1 at separate times

\[\sqrt{2(3)+3}=3 \rightarrow \sqrt{6+3}=3 \rightarrow \sqrt{9}=3 \rightarrow 3=3\]

\[\sqrt{2(-1)+3}=3 \rightarrow \sqrt{-2+3}=3 \rightarrow \sqrt{1}=3 \rightarrow 1\neq3\]

Even thought the solution x = 3 satisfies the equation, x = -1 doesn't so the solutions are extraneous
Therefore lincoln is correct

Answer 37

You understand now how i solved and got the answer?

Answer 38

wait it should be 1 doesn't equal -1 but Lincoln is still correct sorry for the error

Answer 39

im reviewing it!

Answer 40

Ok im see it now

Answer 41

wait in the problem It says is gemme or lincoln correct or are they both correct