Can someone help me solve this equation? Step by Step.

Question

misty1212 · Answer

HI!!

misty1212 · Answer

is it a hard equation ?

airyana1114 · Answer

$$\sqrt{x+6}-4=x$$

misty1212 · Answer

oh no that can't be that hard

airyana1114 · Answer

Can you help me? I've never had this type of equation equaled to x and it sort of makes me nervous. Can you walk me through it? @misty1212

misty1212 · Answer

yeah no problem 
first step is to get the radical by itself on one side of the equation by adding 4 to both sides, i.e. write 
$$\sqrt{x+6}=x+4$$

misty1212 · Answer

now you want to do it the math teacher way , or you want to think what the answer is , or both?

airyana1114 · Answer

I'm not sure I understand what you're asking.

misty1212 · Answer

we can find the answer in our head, but that is probably not what your math teacher wants you to do ,so lets go ahead and solve it the way they do in math class

airyana1114 · Answer

Okay

misty1212 · Answer

i can show you how to get the answer in your head second

airyana1114 · Answer

Alright thanks :)

misty1212 · Answer

you want to get rid of the radical, so square both sides (carefully)

airyana1114 · Answer

How would I square (x+4). Like this? x^2+8

misty1212 · Answer

$$\sqrt{x+6}=x+4$$ square and get $$x+6=(x+4)^2$$ or 
$$x+6=(x+4)(x+4)$$ or $$x+6=x^2+8x+16$$

airyana1114 · Answer

Where did the 8x come from?

misty1212 · Answer

no dear that is not how you square $x+4$
you have to write it as $$x(x+4)(x+4)=x^2+4x+4x+16=x^2+8x+16$$

misty1212 · Answer

it is that old "first outer inner last" stuff

airyana1114 · Answer

Ohhhh, FOIL okay..that makes sense.

misty1212 · Answer

so now we have 
$$x+6=x^2+8x+16$$ which is a quadratic equation we have to solve

misty1212 · Answer

set it equal to zero, get $$x^2+7x+10=0$$ then factor

misty1212 · Answer

$$x^2+7x+10=0\
(x+2)(x+5)=0$$ so 
$$x=-2,x=-5$$ but we are not done yet

misty1212 · Answer

you have to check the answer in the original equation 
not the quadratic equation
if you do, you will see that $x=-2$ works but $x=-5$ does not

airyana1114 · Answer

Making it an extraneous solution correct?

misty1212 · Answer

right , which is a silly way of saying it is not a solution but no matter

airyana1114 · Answer

:) thank you @misty1212

misty1212 · Answer

$$\color\magenta\heartsuit$$