Given the equation:
\[\sqrt{2x+1}=3\] x = 4, solution is extraneous x = 4, solution is not extraneous x = 5, solution is extraneous x = 5, solution is not extraneous
\[\sqrt{x-5}+7=11\] x = 21, solution is extraneous x = 21, solution is not extraneous x = 81, solution is extraneous x = 81, solution is not extraneous
\[-4\sqrt{x-3}=12\] x = 0, solution is not extraneous x = 0, solution is extraneous x = 12, solution is not extraneous x = 12, solution is extraneous
I think the first one is x = 4, solution is not extraneous
I think the second on is x = 81, solution is not extraneous
and I have no idea about the third one
so u need help with only the third one?
Yeah...But am I right about one and two?
@ganeshie8
@madrockz
I really need help
First question correct
Second question, INCORRECT> check again
Yeah! So is it x = 81, solution is extraneous
how did u get x = 81 ?
Umm...I think I Square rooted 9
So I'm guessing its not 81
Second q :- \(\sqrt{x-5}+7=11\) put x = 21 see that it satisfies the equatio
thing inside sqrt becomes 21-5 = 16, so taking sqrt it becoems 4 4+7 = 11
so x = 21 is your solution, and its not extraneous. cuz it satisfied the given equaiton
Wow...I was way off lol. Thanks...Do you understand how to do the third one?
:) Third question :- \(-4\sqrt{x-3}=12 \) divide both sides by -4 \( \sqrt{x-3}=3\) square both sides \( x-3 =9\) \(x = ?\)
Would I add 3 then square...because if I just do 9 I get 81
add 3 both sides and stop there.
oh so 12
x - 3 = 9 x = 12
yes, put x = 12 in given equation, and see if it satisfies or not
tbh, i wont do that cuz i see that sqrt can never become negative. but the given equation requires sqrt to be negative. so the solution x=12 in extraneous. it wont satisfy given equation
see if that makes some sense
Yes defiantly so x = 12, solution is not extraneous
Careful
solution IS extraneous
Oh wrong one....that's the part that keeps messing me up
x = 12, solution is extraneous
yes
IF the solution DOES NOT satisfy given equation, we call it EXTRANEOUS
Okay...I'll remember that! Thanks for helping! ^-^
np :)
Join our real-time social learning platform and learn together with your friends!