Which one of the following is a square root of \(\rm 2a+2\sqrt{a^2 + b^2}\) , where a, b ∈ ℝ?
Finding the Square Root of 2a+2\(\sqrt{a^2 + b^2}\)
The question asks for a square root of the expression \(\rm 2a+2\sqrt{a^2 + b^2}\), where \(a\) and \(b\) are real numbers. We are given options that involve complex numbers.
To determine which option is a square root, we can square each option and see which one results in the original expression \(\rm 2a+2\sqrt{a^2 + b^2}\).
Let's analyze the first option: \(\rm \sqrt{a + ib}+ \sqrt{a - ib}\). We will square this expression:
Let \(X = \sqrt{a + ib}\) and \(Y = \sqrt{a - ib}\). We want to calculate \((X+Y)^2\).
Using the algebraic identity \((X+Y)^2 = X^2 + Y^2 + 2XY\):
\[ \left(\sqrt{a + ib}+ \sqrt{a - ib}\right)^2 = \left(\sqrt{a + ib}\right)^2 + \left(\sqrt{a - ib}\right)^2 + 2\left(\sqrt{a + ib}\right)\left(\sqrt{a - ib}\right) \]
Step 1: Square the terms \(\left(\sqrt{a + ib}\right)^2\) and \(\left(\sqrt{a - ib}\right)^2\).
\[ \left(\sqrt{a + ib}\right)^2 = a + ib \] \[ \left(\sqrt{a - ib}\right)^2 = a - ib \]
Step 2: Multiply the two square root terms \(2\left(\sqrt{a + ib}\right)\left(\sqrt{a - ib}\right)\). We can combine the terms under a single square root because the product of the terms inside the square roots is real (\((a+ib)(a-ib) = a^2 - (ib)^2 = a^2 - i^2 b^2 = a^2 - (-1)b^2 = a^2 + b^2\), and \(a^2+b^2 \ge 0\) for real \(a\) and \(b\)).
\[ 2\left(\sqrt{a + ib}\right)\left(\sqrt{a - ib}\right) = 2\sqrt{(a + ib)(a - ib)} \] \[ = 2\sqrt{a^2 - (ib)^2} \] \[ = 2\sqrt{a^2 - i^2 b^2} \] Since \(i^2 = -1\), we have: \[ = 2\sqrt{a^2 - (-1)b^2} \] \[ = 2\sqrt{a^2 + b^2} \]
Step 3: Combine the results from Step 1 and Step 2.
\[ \left(\sqrt{a + ib}+ \sqrt{a - ib}\right)^2 = (a + ib) + (a - ib) + 2\sqrt{a^2 + b^2} \] \[ = a + ib + a - ib + 2\sqrt{a^2 + b^2} \] \[ = (a + a) + (ib - ib) + 2\sqrt{a^2 + b^2} \] \[ = 2a + 0 + 2\sqrt{a^2 + b^2} \] \[ = 2a + 2\sqrt{a^2 + b^2} \]
The result matches the original expression \(\rm 2a+2\sqrt{a^2 + b^2}\). Therefore, \(\rm \sqrt{a + ib}+ \sqrt{a - ib}\) is a square root of the given expression.
Let's briefly consider the second option, \(\rm \sqrt{a + ib}- \sqrt{a - ib}\). Squaring this gives \(\rm (a+ib) + (a-ib) - 2\sqrt{(a+ib)(a-ib)} = 2a - 2\sqrt{a^2+b^2}\), which is not the target expression.
The third and fourth options are \(\rm 2a + ib\) and \(\rm 2a - ib\). Squaring these would result in expressions involving terms with \(i\) (e.g., \((2a+ib)^2 = 4a^2 - b^2 + 4aib\)), which do not match the form of the original expression \(\rm 2a+2\sqrt{a^2 + b^2}\) which is purely real since \(a\) and \(b\) are real numbers and \(\sqrt{a^2+b^2}\) is a real number.
Thus, the square root of \(\rm 2a+2\sqrt{a^2 + b^2}\) from the given options is \(\rm \sqrt{a + ib}+ \sqrt{a - ib}\).
Revision Table: Key Concepts in Finding Square Roots
| Concept | Description | Relevance to Problem |
|---|---|---|
| Square Root | A number \(x\) is a square root of a number \(y\) if \(x^2 = y\). Every positive real number has two real square roots (one positive, one negative). Complex numbers also have square roots. | We are asked to find a quantity whose square is \(\rm 2a+2\sqrt{a^2 + b^2}\). |
| Complex Numbers | Numbers of the form \(p + iq\), where \(p\) and \(q\) are real numbers and \(i = \sqrt{-1}\). | The options for the square root involve complex numbers. |
| Algebraic Identities | Formulas like \((x+y)^2 = x^2 + y^2 + 2xy\) and \((x-y)^2 = x^2 + y^2 - 2xy\). | Used to expand and simplify the square of the given option. |
| Product of Conjugates | For a complex number \(p+iq\), its conjugate is \(p-iq\). The product is \((p+iq)(p-iq) = p^2 - (iq)^2 = p^2 + q^2\), which is a real number. | Used to simplify \((a+ib)(a-ib)\) inside the square root. |
Additional Information on Square Roots and Complex Numbers
The square root of a non-negative real number \(r\) is usually denoted by \(\sqrt{r}\), which represents the principal (non-negative) square root. However, \(r\) also has \(-\sqrt{r}\) as a square root.
Every non-zero complex number \(z\) has exactly two square roots. If one square root is \(w\), the other is \(-w\).
In this problem, we squared the expression \(\rm \sqrt{a + ib}+ \sqrt{a - ib}\) and got \(\rm 2a+2\sqrt{a^2 + b^2}\). This means \(\rm \sqrt{a + ib}+ \sqrt{a - ib}\) is one of the square roots of \(\rm 2a+2\sqrt{a^2 + b^2}\). The other square root would be \(-\left(\rm \sqrt{a + ib}+ \sqrt{a - ib}\right)\), which was not given as an option.
The expression \(\rm 2a+2\sqrt{a^2 + b^2}\) is always non-negative for real \(a, b\). This is because \(a^2+b^2 \ge 0\), so \(\sqrt{a^2+b^2}\) is a real number \(\ge 0\). The term \(2\sqrt{a^2+b^2}\) is therefore \(\ge 0\). The term \(2a\) can be positive or negative. However, notice the structure: \(\rm 2a+2\sqrt{a^2 + b^2}\) looks like \(\rm x^2\) where \(x = \sqrt{u} + \sqrt{v}\) leading to \(u+v+2\sqrt{uv}\). In our case, if we consider \(u=a+ib\) and \(v=a-ib\), then \(u+v = 2a\) and \(uv = a^2+b^2\), leading to \(2a+2\sqrt{a^2+b^2}\). This confirms that \(\rm \sqrt{a + ib}+ \sqrt{a - ib}\) is a valid form for a square root.
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