How do we solve $x + 3\sqrt{x} − 1/2 = 0$?

How do we solve $x + 3\sqrt{x} − 1/2 = 0$?

Question

1. Use the substitution $\sqrt{x} = y$ (where $y \geq 0$) to find the real root of the equation $x + 3\sqrt{x} − \tfrac{1}{2} = 0.$
2. Find all real roots of the following equations:
3. $x + 10\sqrt{x + 2} − 22 = 0$;
4. $x^2 − 4x + \sqrt{2x^2−8x−3} − 9 = 0$.

Solution

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 $y$): $x+3\sqrt{x}-\frac{1}{2}=\frac{10-3\sqrt{11}}{2}+3\frac{-3+\sqrt{11}}{2}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0.$

---

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:
2. $x^2 − 4x + \sqrt{2x^2−8x−3} − 9 = 0$.

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 $x=2\pm\sqrt{10}$.