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

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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.