# How do you solve #sqrt(4y-9) - sqrt(5y-4) = 1#?

##### 1 Answer

#### Answer:

There are no solutions (Real or Complex), unless we are using the less common definition of

#y = 4-6i#

#### Explanation:

We can attempt to solve the problem as follows:

First add

#sqrt(4y-9) = sqrt(5y-4)+1#

Then square both sides (noting that this may introduce spurious solutions):

#4y-9 = (5y-4)+2sqrt(5y-4)+1#

#=5y-3+2sqrt(5y-4)#

Subtract

#-y-6 = 2sqrt(5y-4)#

Square both sides to get:

#y^2+12y+36 = 4(5y-4) = 20y-16#

Subtract

#y^2-8y+52 = 0#

The discriminant of this quadratic is:

#(-8)^2-(4*1*52) = 64-208 = -144 < 0#

So there are no Real soutions.

**Alternative method**

If we are considering only Real solutions, then we require:

#4y - 9 >= 0# and#5y - 4 >= 0#

in order that the radicands are non-negative.

So

In particular, note that

Hence

So:

**Complex solutions?**

We have already established that any solution

#y^2-8y+52 = 0#

which has discriminant

So the roots of this quadratic are:

#y = (-b+-sqrt(Delta))/(2a) = (8+-12i)/2 = 4+-6i#

If one of these roots is a solution of the original equation, then so is the other, since the original problem had only Real coefficients.

Let

Then:

#sqrt(4y-9) = sqrt(7+24i) = 4+3i#

#sqrt(5y-4) = sqrt(16+30i) = 5+3i#

So:

#sqrt(4y-9) - sqrt(5y - 4) = (4+3i)-(5+3i) = -1 != 1#

So there are no Complex solutions either.

**Principal complications**

My statement "If one of these roots is a solution of the original equation, then so is the other" above is not quite correct. Principal square roots introduce a subtle asymmetry.

Let

If we use the most common definition with

#sqrt(4y-9) = sqrt(7-24i) = 4-3i#

#sqrt(5y-4) = sqrt(16-30i) = 5-3i#

and we find:

#sqrt(4y-9) - sqrt(5y - 4) = (4-3i)-(5-3i) = -1 != 1#

But if we use the less common definition with

#sqrt(4y-9) = sqrt(7-24i) = -4+3i#

#sqrt(5y-4) = sqrt(16-30i) = -5+3i#

and we find:

#sqrt(4y-9) - sqrt(5y-4) = (-4+3i)-(-5+3i) = 1#

a solution!