Can we solve $\sqrt{3-3x} - \sqrt{2-x} = 1$?

Can we solve $\sqrt{3-3x} - \sqrt{2-x} = 1$?

Question

1. Find all pairs of values of $x$ and $y$ for which

$$\begin{equation*} \frac{3x+y+1}{8} = \frac{x-y}{5} = \frac{x^2 - y^2}{5}. \end{equation*}$$

1. Solve the equation

$$\begin{equation*} \sqrt{3-3x} - \sqrt{2-x} = 1. \end{equation*}$$

The expression $\sqrt{u}$ denotes the positive square root of $u$.

Suggestion

I have two equations here with two unknowns. So I seem to be dealing with simultaneous equations, though they have been disguised a little.

I really don’t like all these square roots. Perhaps squaring will help?

Solution

Since

$x^2 - y^2 = (x-y)(x+y)$

, we can write the second equation (after multiplying both sides by

$5$

) as

$$\begin{equation*} x-y = (x-y)(x+y). \end{equation*}$$

So either $x-y=0$, or we can divide by $x-y$ to give $x+y=1$.

Case 1: $x - y = 0$

In this case, the first equation holds if and only if $3x+y+1=0$.

But as $x=y$, this gives $4x+1=0$, so $x=y=-\frac14$ is a solution.

Case 2: $x+y = 1$

In this case, we can write

$y = 1 - x$

and thus the first equation becomes

$$\begin{align*} &&\frac{3x + (1-x) + 1}{8} &= \frac{x - (1-x)}{5} \\ \iff&& \frac{2x + 2}{8} &= \frac{2x - 1}{5} \\ \iff&& 5(x+1) &= 4(2x-1) \\ \iff&& 9 &= 3x \\ \iff&& x &= 3. \end{align*}$$

So $x=3$, $y=-2$ is a second solution.

By squaring both sides, we have

$$\begin{align*} 1 &= \left( \sqrt{3-3x} - \sqrt{2-x} \right)^2 \\ &= (3-3x) - 2\sqrt{3-3x}\sqrt{2-x} + (2-x) \\ &= 5 - 4x - 2\sqrt{3-3x}\sqrt{2-x}. \end{align*}$$

We can rearrange this to put the square roots on one side, giving

$$\begin{equation*} \sqrt{3-3x}\sqrt{2-x} = 2-2x. \end{equation*}$$

By squaring both sides again to get rid of the square root, this implies that

$$\begin{align*} &&(3-3x)(2-x) &= (2-2x)^2 \\ \iff&& 6 - 9x + 3x^2 &= 4 - 8x + 4x^2 \\ \iff&& x^2 + x - 2 &= 0 \\ \iff&& (x+2)(x-1) &= 0 \end{align*}$$

so that $x = 1$ or $x = -2$.

These are only candidate solutions; the original equation implies these are the only solutions possible, but they are only possibilities. Our squaring may have introduced extra ‘solutions’. (For example, if $x=1$, then $x^2=1$, but $x^2=1$ has two solutions, $x=1$ and $x=-1$.) So we need to check whether these candidate solutions are actually solutions of the original equation.

If

$x = 1$

, then

$$\begin{equation*} \sqrt{3-3x} - \sqrt{2-x} = -1 \ne 1, \end{equation*}$$

so this is not a solution to the original problem.

If

$x = -2$

, then

$$\begin{equation*} \sqrt{3-3x} - \sqrt{2-x} = \sqrt{9} - \sqrt{4} = 3 - 2 = 1. \end{equation*}$$

So $x = -2$ is the unique solution to the original problem.