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.?

Snapshot from bookSnapshot from book

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

1 Answer
Jul 2, 2018

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 xes and check the solutions in the original equation.