Question

Solve for x: √(2t-1)+√(3t+3)=5
natalie135....Can you help me again? Please.. Thanks in advance!! (:

Answer 1

OK:
Reduce the number of square roots on the left side by squaring:

\[2+5t+2\sqrt{(2t-1)(3t+3)}=25\]

Isolate the radical:

\[2\sqrt{(2t-1)(3t+3)}=23-5t\]

Raise both sides to the power of two:
\[4(2t-1)(3t+3)=(23-5t)^2\]

Expand:
\[24t^2+12t-12=25t^2-230t+529\]

Combine on one side:
\[t^2-242t+541=0\]

We have to solve this one by completing the square or the quadratic formula:
Complete the square
\[t^2-242t+14641=14100\]
So: \[(t-121)^2=14100\]
OR
plug into the quadratic equation which I dont feel like putting into the equation editor.

And we get:

\[t=121+10\sqrt{141}\]
and \[t=121-10\sqrt{141}\]

When we check the solutions by plugging them back into the original we see that only

\[t=121-10\sqrt{141}\] is correct. So that is the only answer

Answer 2

you have to end up to this answer
242-√56400/2
is your answer the same in this answer??

Answer 3

i got it! Thank you! (: