anonymous
  • anonymous
Solve for x?
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[-4\sqrt{x+9}=20 \]
anonymous
  • anonymous
sqrt(x+9)=-(20/4)
anonymous
  • anonymous
I know how to solve is, I'm just not sure how to get rid of the -4 in the front of the radical.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
you need to remove the square from x+9 the way you do that is by squaring both side and you end up with x+9=-(20/4)^2. then solve from there.
anonymous
  • anonymous
yes that 's it
anonymous
  • anonymous
x+9=-20/4)^2. -20/4^2 = -25 then take -25 subtract 9. you will get the answer of x=-34
anonymous
  • anonymous
it's easy to see that this has NO SOLUTION: (-4)*(-5) = 20 so that means: \(\large -4\sqrt{x+9}=20 \rightarrow \sqrt{x+9}=-5\) \(\large \sqrt{x+9}=-5\) which has no solution because the left side is always positive but the right side is negative
anonymous
  • anonymous
Can't you just square both sides to get rid of the square root, then solve like an equation?
anonymous
  • anonymous
yes you can try to do that... it's one technique you should try to solve it. but why continue when you know this is an impossible equation to solve?
anonymous
  • anonymous
not impossible, no solution... if you do solve it, the solution is an extraneous solution....
anonymous
  • anonymous
If according to your answer it is √x+9 = -5 then you'd square both sides to have: x+9 = -25 subtract 9 from both sides and get -16 as x?
anonymous
  • anonymous
Difficult equation in history
anonymous
  • anonymous
squaring both sides you get positive 25
anonymous
  • anonymous
Oh, that's right thanks so then x+9=25 subtract 9 then positive 16?
anonymous
  • anonymous
yes... now let's see if the solution of x=16 works....
anonymous
  • anonymous
no no no no =-20
anonymous
  • anonymous
that's wrong
anonymous
  • anonymous
\(\large (-4)(\sqrt{x+9})=20 \) \(\large (-4)(\sqrt{\color {red}{16}+9})=20 \) \(\large (-4)(\sqrt{\color {red}{25}})=20 \) \(\large -4 \cdot 5 = 20 \) \(\large -20=20 \) ???? not a true statement
anonymous
  • anonymous
so the solution of x=16 is extraneous and not really a solution.
anonymous
  • anonymous
solution \[\left| -4 \right|\]
anonymous
  • anonymous
suggestion hhhh
anonymous
  • anonymous
@AntarAzri , what's the solution?
anonymous
  • anonymous
\[\left| -4 \right|\]
anonymous
  • anonymous
\(\large (-4)(\sqrt{x+9})=20 \) \(\large (-4)(\sqrt{|-4|+9})=20 \) \(\large (-4)(\sqrt{4+9})=20 \) \(\large (-4)(\sqrt{13})=20 \) x=|-4| doesn't seem to work. That's not a solution.
anonymous
  • anonymous
my oh my I have headaches because of this problem..
anonymous
  • anonymous
no no i dont mean the x |−4|=4
anonymous
  • anonymous
ok. whatever... my answer stands... NO SOLUTION to the equation: \(\large (-4)(\sqrt{x+9})=20 \) \)
anonymous
  • anonymous
\[(\left| -4 \right|)(\sqrt{16+9})=20\]
anonymous
  • anonymous
here's wolframs answer: http://www.wolframalpha.com/input/?i=solve+-4*sqrt%28x%2B9%29%3D20
anonymous
  • anonymous
good
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=solve+|%E2%88%924|*sqrt%28x%2B9%29%3D20
anonymous
  • anonymous
?
anonymous
  • anonymous
if u don't understand my answer and expanation maybe someone else can give u a better answer... i will not argue my point further....
anonymous
  • anonymous
Do we agree that it can not be the root of a negative number gives a positive number ?
anonymous
  • anonymous
I'll see what my teacher says when she grades it. Thanks for your time

Looking for something else?

Not the answer you are looking for? Search for more explanations.