Irrational equation

[trivun]

I've tried dividing both sides with square root of (x-1)(x+3) but I've gotten something very complex. I guess there is some other trick I don't know about.

 

[Dr.Peterson]

I've tried dividing both sides with square root of (x-1)(x+3) but I've gotten something very complex. I guess there is some other trick I don't know about.

I observe that, if we call the first two radicals u and v, the LHS is u + v + 2uv. What is the RHS equal to in terms of u and v? A little work, and you'll find that the whole equation can be reduced to an equation in x = u + v. There are probably other useful tricks, but this worked for me. (The solution turns out to be very simple!)

 

[trivun]

Can you follow this link?
It suggest that you first subtract \(\sqrt{(x-1)(x+3)}\) from both sides and square the whole equation.

Are you suggesting me to do this?

 

[Steven G]

No, bring the whole term including the 2 to the other side. Then square bot sides. Then rearrange and square both sides again.

 

[Dr.Peterson]

View attachment 23276
I observe that, if we call the first two radicals u and v, the LHS is u + v + 2uv. What is the RHS equal to in terms of u and v? A little work, and you'll find that the whole equation can be reduced to an equation in x = u + v. There are probably other useful tricks, but this worked for me. (The solution turns out to be very simple!)

I'd added more about this somewhere, but it seems to have disappeared.

I got u + v + 2uv = 6-u^2-v^2, which can be rearranged to u^2 + 2uv + v^2 + u + v = 6, which then becomes (u+v)^2 + (u+v) - 6 = 0. Solve for (u+v) and you have a simpler radical equation to work on.