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

##### 1 Answer
May 15, 2016

There are no solutions (Real or Complex), unless we are using the less common definition of $A r g \left(z\right) \in \left[0 , 2 \pi\right)$, which yields one solution:

$y = 4 - 6 i$

#### Explanation:

We can attempt to solve the problem as follows:

First add $\sqrt{5 y - 4}$ to both sides to get:

$\sqrt{4 y - 9} = \sqrt{5 y - 4} + 1$

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

$4 y - 9 = \left(5 y - 4\right) + 2 \sqrt{5 y - 4} + 1$

$= 5 y - 3 + 2 \sqrt{5 y - 4}$

Subtract $5 y - 3$ from both ends to get:

$- y - 6 = 2 \sqrt{5 y - 4}$

Square both sides to get:

${y}^{2} + 12 y + 36 = 4 \left(5 y - 4\right) = 20 y - 16$

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

${y}^{2} - 8 y + 52 = 0$

The discriminant of this quadratic is:

${\left(- 8\right)}^{2} - \left(4 \cdot 1 \cdot 52\right) = 64 - 208 = - 144 < 0$

So there are no Real soutions.

$\textcolor{w h i t e}{}$
Alternative method

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

$4 y - 9 \ge 0$ and $5 y - 4 \ge 0$

in order that the radicands are non-negative.

So $y \ge \frac{9}{4}$

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

Hence $\sqrt{5 y - 4} > \sqrt{4 y - 9}$

So: $\sqrt{4 y - 9} - \sqrt{5 y - 4} < 0$ and cannot be equal to $1$.

$\textcolor{w h i t e}{}$
Complex solutions?

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

${y}^{2} - 8 y + 52 = 0$

which has discriminant $- 144 = - {12}^{2}$

So the roots of this quadratic are:

$y = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{8 \pm 12 i}{2} = 4 \pm 6 i$

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 $y = 4 + 6 i$

Then:

$\sqrt{4 y - 9} = \sqrt{7 + 24 i} = 4 + 3 i$

$\sqrt{5 y - 4} = \sqrt{16 + 30 i} = 5 + 3 i$

So:

$\sqrt{4 y - 9} - \sqrt{5 y - 4} = \left(4 + 3 i\right) - \left(5 + 3 i\right) = - 1 \ne 1$

So there are no Complex solutions either.

$\textcolor{w h i t e}{}$
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 $y = 4 - 6 i$

If we use the most common definition with $A r g \left(z\right) \in \left(- \pi , \pi\right]$ then

$\sqrt{4 y - 9} = \sqrt{7 - 24 i} = 4 - 3 i$

$\sqrt{5 y - 4} = \sqrt{16 - 30 i} = 5 - 3 i$

and we find:

$\sqrt{4 y - 9} - \sqrt{5 y - 4} = \left(4 - 3 i\right) - \left(5 - 3 i\right) = - 1 \ne 1$

But if we use the less common definition with $A r g \left(z\right) \in \left[0 , 2 \pi\right)$ then

$\sqrt{4 y - 9} = \sqrt{7 - 24 i} = - 4 + 3 i$

$\sqrt{5 y - 4} = \sqrt{16 - 30 i} = - 5 + 3 i$

and we find:

$\sqrt{4 y - 9} - \sqrt{5 y - 4} = \left(- 4 + 3 i\right) - \left(- 5 + 3 i\right) = 1$

a solution!