How to solve the problem?5^square root(x-3)+7^square root(x^2-8x+15)=2

Before solving the equation, we'll impose the constraints
for the square roots to
exist.

x-3>=0

x>=3

x^2
- 8x+15>=0

We'll determine the roots of the
quadratic:

x1 =
[8+sqrt(64-60)]/2

x1 = 5

x2 =
3

The quadratic is positive over the intervals (-infinite ;
3]U[5 ; +infinite)

Now, we'll re-write the quadratic as a
product of linear factors:

5^square root(x-3)+7^square
root[(x-3)(x-5)]=2

This is a non-standard equation. In this
case, we'll verify first if the solution of the equation is unique. For this reason,
we'll check if the expression from the left side is strictly
increasing.

We notice that the terms of the sum from the
left side are exponentials, which are positive all the time, for any value of x, so the
result of the sum is striclty positive.

The expresison
5^square root(x-3)+7^square root[(x-3)(x-5)]>0 and the solution of the equation
is unique.

Now, we notice that x = 3 is the root of both
radicands from exponents.

We'll put x = 3 and we'll
get:

5^sqrt(3-3) + 7^sqrt(3-3)(3-5) =
2

5^0 + 7^0 = 2

1 + 1 =
2

2 = 2

So, the
only solution of the equation is x = 3.