Question

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

Answer 1

There are no solutions (Real or Complex), unless we are using the less common definition of `Arg(z) in [0, 2pi)`, which yields one solution:

`y = 4-6i`

Explanation:

We can attempt to solve the problem as follows:

First add `sqrt(5y-4)` to both sides to get:

`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 `5y-3` from both ends to get:

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

Square both sides to get:

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

Subtract `20y-16` from both ends to get:

`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.

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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 `y >= 9/4`

In particular, note that `5y - 4 > 4y - 9` since both `-4 > -9` and `5y > 4y`.

Hence `sqrt(5y - 4) > sqrt(4y - 9)`

So: `sqrt(4y -9) - sqrt(5y-4) < 0` and cannot be equal to `1`.

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Complex solutions?

We have already established that any solution `y` must satisfy:

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

which has discriminant `-144 = -12^2`

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 `y = 4+6i`

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.

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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 `y = 4 - 6i`

If we use the most common definition with `Arg(z) in (-pi, pi]` then

`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 `Arg(z) in [0, 2pi)` then

`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!