Solving sqrt (Y+6) - sqrt (3y) = 3

  • Thread starter Thread starter perpetual-burn
  • Start date Start date
AI Thread Summary
To solve the equation sqrt(Y+6) - sqrt(3y) = 3, it is necessary to square both sides of the equation twice. The first squaring isolates one square root, leading to the equation y + 6 + 3y - 2sqrt((y+6)(3y)) = 9. After isolating the remaining square root, squaring again will yield a quadratic equation. It is important to check that the solutions obtained are greater than zero to ensure they are valid. This method will help in accurately completing the homework.
perpetual-burn
Messages
7
Reaction score
0
Hey guys,
I was wondering how to solve this equation. I tihnk I got it, but I want to make sure so I don't go on doing all of my homework wrong.

Question: sqrt (Y+6) - sqrt (3y) = 3

I forget if I have to sqaure each separate or the whole left side as FOIL. Please let me knwo asap so I can finish homeowkr :) Again, thanks so much!

-Phil
 
Physics news on Phys.org
U'll have to square twice.And each time separate a sqrt in one side of the equation.

\sqrt{y+6}-\sqrt{3y}=3\Rightarrow y+6+3y-2\sqrt{\left(y+6\right)3y}=9

and then isolate in one member of the equation that sqrt and then square again.

Daniel.
 
Ok, waht i tohught :) Thanks Dan.
 
I hope u can solve the quadratic that will follow.And to check whether the solutions u get are larger than 0...

Daniel.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Solving a Puzzling Physics Problem: ##15 \pm 3\sqrt{15}##
Replies
17
Views
2K
Are the 2nd and 3rd problems in this video correctly solved? (charged dipole and organ pipe problems)
Replies
3
Views
980
Why Does Solving y+3=3√(y+7) Yield Extraneous Solutions?
Replies
8
Views
2K
Time period of SHM & Energy conservation in pulley-spring-block system
Replies
24
Views
3K
Find v_1 for projectile using start position, final position and angle
Replies
10
Views
2K
To find the resultant of forces on a lamina
Replies
5
Views
3K
Evaluate ##\dfrac {x^\frac{3}{2}+xy}{xy-y^3}-\dfrac {\sqrt x}{\sqrt x -y}##
Replies
13
Views
2K
Help Needed: Solving y = A sin({2*10^6 ({\pi / 3 -\pi t }) + \phi })
Replies
2
Views
954
Closest distance of approach between 2 charged particles
Replies
5
Views
1K
How to Solve the Motion of a Chain Falling Off a Table?
Replies
6
Views
3K

Hot Threads

Recent Insights

Back
Top