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

giorgiana1976 | Student

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.