Japanese | A Nice Math Olympiad Algebra Problem | A-MATHS

 Japanese | A Nice Math Olympiad Algebra Problem | A-MATHS

The given equation is:
\sqrt{x} + \sqrt{-x} = 4

Let’s solve this step by step.

Step 1: Understand the Domains

First, note that the square root function \(\sqrt{y}\) is only defined for non-negative values of \(y\). Therefore, both \(x\) and \(-x\) must be non-negative.

This means \(x \geq 0\) and \(-x \geq 0\).

For \(-x \geq 0\), \(x \leq 0\).

Combining these two conditions, we get:
x = 0

However, let’s check if \(x = 0\) is a solution.

Step 2: Check \(x = 0\)

Substituting \(x = 0\) into the equation:

\sqrt{0} + \sqrt{-0} = 0 + 0 = 0

This does not satisfy the equation \(\sqrt{x} + \sqrt{-x} = 4\).

So, \(x = 0\) is not a solution. Let’s consider if the equation allows for complex numbers.

Step 3: Consider Complex Solutions

Let’s rewrite the equation:
\sqrt{x} + \sqrt{-x} = 4

For complex numbers, we can express \(x\) as:

x = re^{i\theta}

where \(r \geq 0\) is the modulus and \(\theta\) is the argument.

Then, \(\sqrt{x} = \sqrt{r}e^{i\theta/2}\) and \(\sqrt{-x} = \sqrt{r}e^{i(\theta/2 + \pi/2)}\).

Substituting into the original equation:

\[
\sqrt{r}e^{i\theta/2} + \sqrt{r}e^{i(\theta/2 + \pi/2)} = 4
\]

Step 4: Separate Real and Imaginary Parts

For the sum to be real (since 4 is real), the imaginary parts must cancel out. Let’s choose \(\theta = \pi\), which means \(x = -r\). This choice simplifies to:
\sqrt{-r} + \sqrt{r} = 4

Since \(r\) is non-negative, let \(r = a²\) (where \(a \geq 0\)). Then:
\sqrt{a²} + \sqrt{-a²} = 4

This means:
a + ai = 4

Step 5: Solve for \(a\)

For the imaginary part to be zero:
ai = 0

Thus, \(a = 0\). But substituting \(a = 0\):
0 + 0 = 4

This contradiction implies there might be a mistake in our approach. Let’s consider the initial assumption.

Step 6: Revisit Real Numbers

Let’s try another approach assuming \(x\) could have a specific complex form. If we consider \(x = -k²\):
\sqrt{-k²} = ki

Thus:
ki + k = 4

Separating real and imaginary:
k(1+i) = 4

Then:
k = \frac{4}{1+i}

Multiplying by the conjugate:
k = \frac{4(1-i)}{1+1} = 2(1-i)

So, \(x = -k²\):

\[
x = -[2(1-i)]² = -4(1–2i+i²) = -4(1–2i-1) = 8i
\]

Finally, checking:
\sqrt{8i} + \sqrt{-8i} = \sqrt{8}i + \sqrt{8}(-i) = 0

which doesn’t satisfy the equation.

Conclusion

Therefore, the original equation \(\sqrt{x} + \sqrt{-x} = 4\) has no real solutions, and attempts to solve it using complex numbers lead to contradictions or invalid results. Thus, we conclude there are no solutions for the given equation.