Question

How do I know if this solution is extraneous? Please help. sqrt {x+7}-x=1

Answer 1

\[\sqrt{x+7}-x=1\]

Answer 2

\[\begin{array}{rcl}
\sqrt{x+7}-x&=&1\\
\sqrt{x+7}&=&1+x\\
x+7&=&x^2+2x+1\\
x^2-x-6&=&0\\
(x-3)(x+2)&=&0\\
x-3=0&or&x+2=0\\
x=3&or&x=-2
\end{array}\]

Answer 3

Substitute both solutions back to the equation and see if the equation is valid

Answer 4

If the equation is not valid, then that solution is extraneous.

Answer 5

i think...
sqrt {x+7}-x = 1
sqrt {x+7} = 1+ x
(sqrt {x+7})^2 = ( 1+ x )^2
x+7 = 1 + 2x + x^2
0 = 1 + 2x + x^2 - x - 7
0 = -6 + x + x^2
0 = (x-2)(x+3)
so x = 2 or x = -3
pick which is right by substitution

Answer 6

*sigh*... u types whay quicker @kc_kennylau

Answer 7

;D

Answer 8

@Misskatierae Please give @Jack1 a medal because he answered first, I just typed quicker than him.

Answer 9

nah, fastest finger first @Misskatierae , give it to @kc_kennylau

Answer 10

@Misskatierae yep medal him you've made the correct decision :)

Answer 11

lol, i think we're spamming u're Q now @Misskatierae , sorry ;D
cheers kc

Answer 12

How about whoever can explain how they got the answer gets a medal? Because I'm still confused haha

Answer 13

u go @kc_kennylau , it'll be quicker ;D

Answer 14

Haha!! @kc_kennylau can you please help?

Answer 15

Wait my answer was wrong Jack's answer was correct

Answer 16

@Jack1 can you please help? lol

Answer 17

When x=2,
L.H.S.
=sqrt(2+7)-2
=sqrt9-2
=3-2
=1
R.H.S.
=1
\(\therefore\)x=2 is not extraneous

Answer 18

When x=-3,
L.H.S.
=sqrt(-3+7)-2
=sqrt4-2
=2-2
=0
R.H.S.
=1
\(\therefore\)x=-3 is extraneous.

Answer 19

@Misskatierae please give @Jack1 a medal since he answered correctly :)

Answer 20

hahaha, most gentlemanly answer string ever ;D

Answer 21

@Jack1 yep I agree xD

Answer 22

wait, does that all make sense now @Misskatierae ?

sidebar: (attached)

Answer 23

nope @Jack1

Answer 24

ok... which part are you getting lost at, will slow it down and go into detail

Answer 25

I don't get any of it haha @Jack1 Thank you both so much for helping though!

Answer 26

Complete solution:
\[\begin{array}{rcl}
\sqrt{x+7}-x&=&1\\
\sqrt{x+7}&=&x+1\\
x+7&=&x^2+2x+1\\
x^2+x-6&=&0\\
(x+3)(x-2)&=&0\\
x+3=0&\mbox{or}&x-2=0\\
x=-3&\mbox{or}&x=2
\end{array}\]When x=-3,
\[L.H.S.\\=\sqrt{-3+7}-(-3)\\=\sqrt4+3\\=7\\R.H.S.\\=1\\\because L.H.S.\ne R.H.S.\\\therefore x=-\mbox{3 is extraneous.}\]When x=2,
\[L.H.S.\\=\sqrt{2+7}-2\\=\sqrt9-2\\=3-2\\=1\\R.H.S.\\=1\\\because L.H.S.=R.H.S.\\\therefore x=\mbox{2 is not extraneous.}\]

Answer 27

Thank you soo much @kc_kennylau!!

Answer 28

nice use of LaTex too dude @kc_kennylau

Answer 29

lolz thanks @Jack1 :D