**1. Unique Solution: For which real numbers, a, does the equation a 3**^{x} + 3^{-x} = 3

have a unique solution?

a=9/4 or a≤0.

Whenever I see the words "unique solution" I start looking for a quadratic to solve, knowing I can throw away everything except the

discriminant, which I set to zero and I'm done. Sure enough, there's a quadratic here, in 3^{x}, which can be easily solved. As long as a is not zero:

a 3^{x} + 1/3^{x} = 3

a 3^{2x} - 3*3^{x} + 1 = 0

3^{x} = (3±sqrt(9-4a))/(2a)

Now, this equation has a unique solution in 3^{x} when a=9/4. But what if a=0? Then we have 3^{-x}=3, which has a unique solution. So a=9/4 or a=0, and we're done, right?

Not quite. 3^{x} must be positive (if x is real) so other values of "a" might exist that give two different roots of the quadratic such that one of the roots must be thrown away, leaving a unique solution.

This can only happen if one root is positive and the other is not positive.

There is no value of "a" that can makes a root zero, because of the constant term, so that leaves only the possibility that two non-zero roots have opposite sign, which can only happen if sqrt(9-4a)>3, which can only happen if a<0.

So the final answer is this: a=9/4 or a≤0.

**2. Two Solutions: For what values of m does sqrt(x-5)=mx+2 have two solutions?**

Before we begin, we should note that m can't be 0, because sqrt(x-5)=2 has

only one solution.

First, square both sides to get a quadratic equation:

sqrt(x-5)=mx+2

x-5=(mx+2)^{2}

x-5=m^{2}x^{2} + 4mx + 4

m^{2}x^{2} + (4m-1)x + 9 = 0 (eq. A1)

Now, find the discriminant to get the range of values of m that gives you two solutions:

(4m-1)^{2}-36(m^{2}) > 0

16m^{2}-8m+1-36m^{2} > 0

20m^{2}+8m-1 < 0

(10m-1)(2m+1) < 0

-1/2 < m < 1/10 and m ≠ 0

But we're not done yet, because we need to test these values of m to make sure they really do yield two solutions. You see, when we squared both sides of the equation, we might have introduced spurious roots. Let x_{1},x_{2 }be the two roots of the original equation for a given value of m, with x_{1}< x_{2}. From the quadratic formula we see the roots are:

x_{1} = (-4m+1 - sqrt(-20m^{2}-8m+1))/(2m^{2}), and

x_{2} = (-4m+1 + sqrt(-20m^{2}-8m+1))/(2m^{2}).

Both sides of the original equation must be positive for both roots. In particular, the larger root, x_{2}, must be such that mx_{2}+2

is greater than zero.

mx_{2}+2 > 0

m(-4m+1 + sqrt(-20m^{2}-8m+1))/(2m^{2})+2 > 0

(1 + sqrt(-20m^{2}-8m+1))/(2m) > 0

m > 0

So the final answer is 0 < m < 1/10