Algebra 2 Help Please??

Question

anonymous · Answer

$$\sqrt{x+10} -4=x$$

anonymous · Answer

Identify all of the solutions to that equation

anonymous · Answer

x = –6

 x = –1

 x = –6 and x = –1

 None of the above

debbieg · Answer

Isolate the sq. root expression on one side.

Then square both sides, and you won't have a sq root anymore. :)

Solve the equation as usual (it will be a quadratic)

CHECK each solution back in the original (not unusual to get extraneous solutions, when you start squaring sides of an equation!)  :)

anonymous · Answer

$$\sqrt{x+10}=x+4$$
$$x+10=x ^{2}+16$$
x=-6

anonymous · Answer

right?

anonymous · Answer

x =6 and x =1 are your solutions.

debbieg · Answer

no....

(x+4)^2 is not = x^2 + 16

anonymous · Answer

yeah, wouldn't you distribute the exponent?

debbieg · Answer

I'm not sure what @soda is talking about. Those are most certainly NOT your solutions (but shouldn't be giving answers anyway, the idea is to learn something).

debbieg · Answer

Noooooo no no no. :) That isn't how you multiply binomials.

FOIL!! The square of a binomial is just a particular FOIL.

$(a + b)^2=(a +b)(a+b)=a^2+2ab+b^2$

anonymous · Answer

Oh shoot! I forgot about that!! Thanks :)

anonymous · Answer

x= -6 and = -1, you want a cookie @DebbieG ...

debbieg · Answer

I'm not sure what you mean, @soda. No need to be snippy, I'm just making sure that the asker doesn't take the wrong answers you gave and think they are correct.

And, No, that's not the correct answer either.

anonymous · Answer

It is -1 -6.. I checked them after using foil and it worked

debbieg · Answer

Those are the solutions to the quadratic, which you got after squaring both sides of the original equation.

Did you try them IN THE ORIGINAL EQUATION?

debbieg · Answer

$\sqrt{-6+10}-4=\sqrt{4}-4=2-4=-2$

So, is $\sqrt{x+10}-4=x$, if x=-6?

debbieg · Answer

Like I said above:

CHECK each solution back in the original (not unusual to get extraneous solutions, when you start squaring sides of an equation!) :)

anonymous · Answer

Goodness gracious @DebbieG

anonymous · Answer

@soda stop

anonymous · Answer

Okay Debbie, I get what you're saying now! I would've never thought of that! Thanks so much!

debbieg · Answer

You're very welcome, happy to help. :)