Question

mitosuki Neophyte

solve √x - √5 + 7 = x (The radical is suppose to go over the whole x - 5)
a.x = –6
b.x = –9
c.x = –6 and x = –9
d.None of the above

Answer 1

Write \[\sqrt{x-5} = x - 7\]
and now square both sides, giving you ....

Answer 2

\[\sqrt{x}-\sqrt{5} \neq \sqrt{x-5}\]

Answer 3

but i assume you meant \[\sqrt{x-5}\]

Answer 4

yes that is what I meant

Answer 5

so what happens after completing the step james mentioned

Answer 6

Ok the answer I'm getting is x = 6 and x = 9 but only x=9 works. But none of the answers are there. So it would just be none of the above right? I want to make I'm not doing anything wrong.

Answer 7

\[(\sqrt{x-5})^2=(x-7)^2\] is this what he means by sqaring both sides?
if so, then if you square the square root of something what do you get? that something right?
so
we have
\[x-5=(x-7)^2\]

Answer 8

\[x-5=(x-7)(x-7)\]
\[x-5=x(x-7)-7(x-7)\]
\[x-5=x^2-7x-7x+49\]
\[x-5=x^2-14x+49\]
put everything on one side by subtracting x on both sides and by adding 5 on both sides
\[0=x^2-14x-x+49+5\]

Answer 9

\[0=x^2-15x+54\]
so we need to find two factors of 54 that have product 54 and have sum -15
54=-6(-9)
-15=-6-9
0=(x-6)(x-9)
x=6 or x=9 or is it neither?
check both answers to see
the oringral equation was
\[\sqrt{x-5}+7=x\]
what happens when you plug in 6
1+7=6 noy true
6 is not a solution
what happens when you plug in 9
2+7=9 is a true statement
x=9 is only solution

Answer 10

sigh. I hate it when I can't add.

Answer 11

the only mistake i think i see is i don't know how to spell oringal

Answer 12

adding sucks

Answer 13

lol well thank you guys very much for the help :)

Answer 14

oops another mistake not noy but not lol

Answer 15

1+7=6 not true*

Answer 16

Oki doki thank you

Answer 17

np

Answer 18

oh yeah good job mitosuki
this is what you said you got right?
you deserve a medal :)

Answer 19

haha yes I did thank you again C
:

Answer 20

*C: