how do i solve for x in
√2x-4) - (√x-5) = 3

Question

how do i solve for x in 
√2x-4) - (√x-5) = 3

jamesj · Answer

$$ \sqrt{2x - 4} - \sqrt{x-5} = 3 $$
Yes?

anonymous · Answer

yes

anonymous · Answer

HI JAMES!

jamesj · Answer

Write u = sqrt(x-5).  Then x = u^2 + 5 and sqrt(2x - 4) = sqrt(2u^2 + 6)

Hence we can rewrite the equation above as 

sqrt(2u^2 + 6) = u + 3

Now square both sides:

2u^2 + 6 = u^2 + 6u + 9

Hence u^2 - 6u - 3 = 0

Solving for u, u = 3 +- 2sqrt(3) , or u^2 = 21 + 12sqrt(3) ONLY, because the other solution is negative

Thus x = u^2 + 5 = 26 + 12sqrt(3)

anonymous · Answer

the answers are 6, and 30

jamesj · Answer

No it's not.  For example, substitute 6 back into the equation and you get

sqrt(12 - 4) - sqrt(1) = sqrt(8) - 1 which does not equal 3.

jamesj · Answer

However, 6 and 30 are solutions of 

sqrt(2x + 4) - sqrt(x - 5) = 3.

Notice the change from - to + in the first sqrt.

!!!!!!!

jamesj · Answer

You can show it from scratch by following the procedure I just used for your original version of the problem.

I think I deserve a 'good answer' for that!

anonymous · Answer

so can i solve sqrt(2x + 4) - sqrt(x - 5) = 3.

jamesj · Answer

Yes, make the same substitution I did for u and follow it through, step by step.

What grade are you in?

anonymous · Answer

Thank you.

anonymous · Answer

Hi james !!

anonymous · Answer

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