Question

Please can you solve the problem on an equation in the real number system given in the image below and also tell the sequence to tackle such problems.?

1. No Solution.
2. Exactly two distinct solutions
3. Exactly 4 distinct solutions.
4. Infinitely many solutions

Answer 1

`x=10`

Explanation:

Since `AAx in RR`
`=>`
`x-1>=0`
`and`
`x+3-4sqrt(x-1)>=0`
`and`
`x+8-6sqrt(x-1)>=0`
`=>`
`x>=1` and `x>=5` and `x>=10`
`=>`
`x>=10`

let try then `x=10`:
`sqrt(10+3-4sqrt(10-1))+sqrt(10+8-6sqrt(10-1))=sqrt(13-12)+0=sqrt(1)=1`
so it's not D.
Now try `x=17`
`sqrt(17+3-4sqrt(17-1))+sqrt(17+8-6sqrt(17-1))=sqrt(20-16)+sqrt(25-24)=sqrt(4)+sqrt(1)=2+1=3!=1`
Now try `x=26`
`sqrt(26+3-4sqrt(26-1))+sqrt(26+8-6sqrt(26-1))=sqrt(29-20)+sqrt(34-30)=sqrt(9)+sqrt(4)=3+2=5!=1`
`...`
We can see that when we will take more `x_(k+1)>x_(k)` where `x_k={k^2+1|k in ZZ_(>=3)}`
That to say `{x_k}_(k=3)^oo`
will give us a solution in `ZZ`. both functions are motion-up so the solutions will be bigger than 1.

So I think it must be only 1 solution correct.

---

Alternative way is this:

`sqrt(x+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x-1))=1`

`a^2=b^2 iff a=b or a=-b`

Given we're "living" in `RR`, we know that both `a` and `b` are positive (`a=sqrt(y_1)+sqrt(y_2)>=0` and `b=1>0`):

`(sqrt(x+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x-1)))^2=(1)^2`
`=>`
`x+3-4sqrt(x-1)+x+8-6sqrt(x-1)+2sqrt(x+3-4sqrt(x-1))sqrt(x+8-6sqrt(x-1))=1`
`=>`
`2x+11-10sqrt(x-1)+2sqrt((x+3-4sqrt(x-1))(x+8-6sqrt(x-1)))=1`
`=>`
`-10sqrt(x-1)+2sqrt(...)=-10-2x`
`=>`
`(-10sqrt(x-1)+2sqrt(...))^2=(-10-2x)^2`
`...`
you need to repeat the idea again and again until the "`sqrt`" sign disappears. Than you may get the `x`es and check the solutions in the original equation.