1. Use the substitution \(\sqrt{x} = y\) (where \(y \geq 0\)) to find the real root of the equation \[x + 3\sqrt{x} − \tfrac{1}{2} = 0.\]

Why must we have \(y\geq 0\)? It is because when we write \(\sqrt{x}\), we specifically mean the positive square root (or zero if \(x=0\)).

So when we solve the quadratic in \(y\), we must be careful to only consider positive roots of that quadratic.

As the question suggests, we rewrite the equation with \(y\) for \(\sqrt{x}\), and therefore \(y^2=x\) so \[y^2+3y-\tfrac{1}{2}=0.\]

We solve this by completing the square: \[\begin{align*} &&&\left(y+\frac{3}{2}\right)^2-\frac{9}{4}-\frac{1}{2}=0&&\qquad \\ \iff &&&\qquad \left(y+\frac{3}{2}\right)^2=\frac{11}{4} \\ \iff &&&\qquad y+\frac{3}{2}=\pm \frac{\sqrt{11}}{2} \\ \iff &&&\qquad y=\frac{-3\pm\sqrt{11}}{2}. \end{align*}\]

We could also solve this equation using the quadratic formula or any other method you know.

We need to be a bit careful here—recall that the question told us that we must have \(y\geq 0\). So we take \[y=\frac{-3+\sqrt{11}}{2},\] and so we have \[x=y^2=\frac{10-3\sqrt{11}}{2}.\]

We can check that this satisfies the original equation (remembering we know what \(\sqrt{x}\) is since this is the value we found for \(y\)): \[x+3\sqrt{x}-\frac{1}{2}=\frac{10-3\sqrt{11}}{2}+3\frac{-3+\sqrt{11}}{2}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0.\]

  1. Find all real roots of the following equations:

    1. \(x + 10\sqrt{x + 2} − 22 = 0\);

A good general tip: if you’re given a suggestion for the first part of the question, and you can’t immediately see how to solve the next part, try using a similar method to that suggested in the suggestion.

You may have to modify the suggestion a little, though!

Following the method for the first question, we try the substitution \(y=\sqrt{x+2}\), again for \(y\geq 0\). This time, we have \(y^2=x+2\), or \(x=y^2-2\), so we are now solving the equation \[y^2+10y-24=0.\] This equation can be factorised as \[y^2+10y-24=(y+12)(y-2)=0.\] Again, we can disregard the solution \(y=-12\), since \(y \geq 0\), and see that \(y=2\). This implies that \(x=2\) is the only possible solution. We check whether this is a valid solution: \[x+10\sqrt{x+2}-22=2+10(2)-22=0. \quad \checkmark\]

  1. Find all real roots of the following equations:

    1. \(x^2 − 4x + \sqrt{2x^2−8x−3} − 9 = 0\).

Let’s try the substitution \(y=\sqrt{2x^2-8x-3}\). We can rearrange this to find that \(x^2-4x=(y^2+3)/2\), and so we this time study the equation \[\frac{y^2+3}{2}+y-9=0,\] or equivalently, \[y^2+2y-15=0.\]

We can again factorise to find that \[y^2+2y-15=(y+5)(y-3)=0.\] We must again discard the negative solution, and consider \(y=3\). To find \(x\), we must now solve \(x^2-4x=(3^2+3)/2=6\), so \(x^2-4x-6=0\). Applying the quadratic formula, we deduce that \[x=2\pm\sqrt{10}.\]

When squaring to solve equations, we may gain extra ‘solutions’ that don’t satisfy the original equation, so we must check them.

But in this case, we have \[(x^2 − 4x) + \sqrt{2x^2−8x−3} − 9= 6+3-9=0,\] so the two solutions are indeed \(x=2\pm\sqrt{10}\).